# On the number of total prime factors of an odd perfect number

Joshua Zelinsky

###### Abstract

Let $N$ be an odd perfect number. Let $\omega(N)$ be the number of distinct
prime factors of $N$ and let $\Omega(N)$ be the total number of prime factors
of $N$. We prove that if $(3,N)=1$, then
$\frac{302}{113}\omega-\frac{286}{113}\leq\Omega.$ If $3|N$, then
$\frac{66}{25}\omega-5\leq\Omega.$ This is an improvement on similar prior
results by the author which was an improvement of a result of Ochem and Rao.
We also establish new lower bounds on $\omega(N)$ in terms of the smallest
prime factor of $N$ and establish new lower bounds on $N$ in terms of its
smallest prime factor.

Iowa State University

<EMAIL_ADDRESS>
Recall that a positive integer $N$ is said to be perfect if the sum of $N$’s
proper divisors add up to the number or equivalently that $\sigma(N)=2N$.

It is currently unknown whether there are any odd perfect numbers. Let $N$ be
an odd perfect number. Ochem and Rao[14] have proved that $N$ must satisfy

$\Omega(N)\geq\frac{18\omega(N)-31}{7}$ (1)

and

$\Omega(N)\geq 2\omega(n)+51.$ (2)

Note that Ochem and Rao’s second inequality is stronger than the first as long
as $\omega(N)\leq 81$. Nielsen [12] has shown that $\omega(n)\geq 10$. In a
previous paper [23], the author improved on Ochem and Rao’s result, showing
that:

######  Theorem 1.

If $N$ is an odd perfect number, with $3\not|N$, then

$\Omega(N)\geq\frac{8}{3}\omega(N)-\frac{7}{3}.$ (3)

If $N$ is an odd perfect number, with $3|N$, then

$\Omega(N)\geq\frac{21}{8}\omega(N)-\frac{39}{8}.$ (4)

In this paper we prove

######  Theorem 2.

If $3\not|N$, then

$\frac{302}{113}\omega-\frac{286}{113}\leq\Omega.$ (5)

If $3|N$, then

$\frac{66}{25}\omega-5\leq\Omega.$ (6)

Note that while Inequality 6 is always better than Inequality 4, Inequality 5
is only better than Inequality 3 when $\omega\geq 34$.

Note that the worst case of the above is when $3|N$, and so we have

######  Corollary 3.

If $N$ is an odd perfect number then

$\Omega(N)\geq\frac{66}{25}\omega-5\leq\Omega.$

Note that Kevin Hare [17] has shown that in general any odd perfect number
must satisfy $\Omega\geq 75$, while [13] has improved this to $\Omega\geq
101$. This paper will contain seven sections. The first section contains
various results we will need to prove Theorem 2. The second section contains
the proof of Theorem 2 when $3|N$. The third section contains the proof when
$(3,N)=1$. The fourth section improves on the known lower bound of $\Omega$ in
terms of the smallest prime factor of $N$. This is essentially a small
improvement of existing results although new questions are raised based on
some aspects of the methods used. The fifth section combines the ideas of the
previous sections to improve lower bounds for $N$ in terms of its smallest
prime factor. The sixth section discusses a new way of measuring how strong a
statement is about odd perfect numbers and evaluates the Ochem and Rao type
bounds in that context. The seventh section discusses various related open
problems that are naturally connected to improving these results. We will use
the following notation: $N$ will be an odd perfect number. We will write
$\Omega$ for $\Omega(N)$ and write $\omega$ for $\omega(N)$. We recall Euler’s
classical theorem on odd perfect numbers. Euler proved that $N$ must have the
form $N=q^{e}m^{2}$ where $q$ is a prime such that $q\equiv e\equiv 1$ (mod 4)
and $(q,m)=1$. Traditionally $q$ is called the special prime.111Some authors
call $q$ the “Euler prime.” A better name in fact would be the Cartesian prime
since prior to Euler’s result Descartes proved that an odd perfect number
needed to have exactly one prime factor raised to an odd power. In any event,
special prime avoids any issues of priority. Note that from Euler’s result one
immediately has $\Omega\geq 2\omega-1$. Essentially all improvements on Ochem-
Rao type inequalities can be thought of as improving on the bound one has from
Euler’s theorem. For the remainder of this paper we will assume that $N$ is an
odd perfect with $q$, $e$ and $m$ given as above. The basic method of this
paper is the same as that used in Ochem and Rao’s result. The essential
observation is that if $N$ is an odd perfect number with a prime $p$ raised to
just the second power, then for each such prime $p$ you have $p^{2}+p+1|n$. It
follows from quadratic reciprocity that if $q$ is a prime and $q|p^{2}+p+1$,
then $q$ is either equal to 3 or is 1 mod 3. Now assume that $N$ has many such
primes $p$. If $p^{2}+p+1$ is often divisible by 3, then $N$ will be divisible
by a large power of $3$. If q is not 3 in some instance, than with the
exception of the one prime in an OPN which may be raised to the first power,
one either has $q^{2}||N$ , which gives a new 3 in the factorization since
$q\equiv 1$ (mod 3), and so $3|q^{2}+q+1$, or one at least has $q^{4}|N$. So
the key then is that if one has many numbers of the form $p^{2}+p+1$ showing
that there isn’t much room for a lot of primes repeated exactly twice. And if
one has most primes raised to a higher power then one gets repeated prime
factors from those primes. This idea is made rigorous through a system of
linear inequalities and optimizing that set of inequalities then gives the
type of result. While this paper is substantially longer than the previous
paper by this author or Ochem and Rao’s paper, the basic method remains the
same. The improvements in the case of $3|N$ are essentially straightforward
and represent a small improvement of the technique using some new minor number
theoretic results to get additional inequalities in the system used. The case
of $(3,N)=1$ involves three major new ingredients. Our first new ingredient is
the notion of a triple threat. Define a _triple threat_ , to be a quadruplet
of four odd primes $x,a,b,c$ such that

$\sigma(x^{2})=x^{2}+x+1=\sigma(a^{2})(\sigma(b^{2})\sigma(c^{2})=(a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1)$

and where $a^{2}+a+1$, $b^{2}+b+1$, and $c^{2}+c+1$ are all prime. The primary
obstruction to improving the $8/3$ bound in Theorem 4 arose from the fact that
we could not rule out the existence of odd perfect numbers with many primes
acting triple threats. Consider a triple threat $(x,a,b,c)$ where where
$x\equiv a\equiv 1$ (mod 5). We show that no such triple threats exist with
this property. This is not by itself sufficient to improve the bound. Our
second new ingredient is that if $p^{4}||N$, then every prime divisor of
$\sigma(p^{4})$ is either equal to 5 or is $1$ (mod 5). Our third new
ingredient is the observation that we have the following miraculous
factorization:222This observation seems to have been first noted explicitly in
the literature in [4]. If $f(x)=x^{2}+x+1$, and $g(x)=x^{4}+x^{3}+x^{2}+x+1$,
then

$f(g(x))=(x^{2}-x+1)(x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3).$

This means that if we have a prime $m$ such that $m^{4}||N$, and
$\sigma(m^{4})=p$ is itself prime then we can substantially restrict what
$\sigma(p^{2})$ looks like. We can combine the second and third ingredients to
can use this to guarantee that there are either many primes $p$ where
$p^{2}||N$ and $p\equiv 1$ (mod 5), $N$ is divisible by a large power of $5$,
$N$ has many prime factors raised to at least the sixth power, or $N$ has many
prime factors of a very special form arising from the third ingredient
factorization. Note that all three of these three ingredients are necessary
for our improvement. Any two of them will not by themselves give rise to an
improvement beyond the $8/3$ bound.

##  1 Foundations

This section contains various lemmata we will need for the main results. We
will assume some basic familiarity with the literature on perfect numbers but
will recall some basic facts here. For some early history on this matter see
[5].
In general, while the Greeks originally defined perfect numbers in terms of
the sum of the proper divisors, it is more natural to define a number as
perfect as $n$ satisfying $\sigma(n)=2n$ where $\sigma(n)$ is the sum of all
the positive divisors of $n$. Much of the study of perfect numbers relies on
the nice fact that $\sigma(n)$ is a multiplicative function. Recall that a
number is said to be _abundant_ if $\sigma(n)>2n$. A number is said to be
_deficient_ if $\sigma(2n)<n$. We will set $h(n)=\frac{\sigma(n)}{n}$. Note
that $h(n)$ is called both the _abundancy_ of $n$ or the _index_ of $n$ in the
literature. We have that

$h(n)=\frac{\sum_{d|n}d}{n}=\sum_{d|n}\frac{d}{n}=\sum_{d|n}\frac{1}{d}.$ (7)

From 7 we have that if $a>1$, then we have $h(an)>h(n)$. In particular, any
multiple of an abundant number is itself abundant, and thus no perfect number
can be divisible by an abundant number. Almost all work on odd perfect numbers
relies on the fundamental observation that we can bootstrap from knowing that
a specific prime power divides $N$ to get that other prime powers divide $N$.
For example, if $3^{2}||N$, then $\sigma(3^{2})=13$ must divide $N$. In
general, if $p^{k}||N$ then we must have $\sigma(p^{k})=1+p+p^{2}\cdots
p^{k}|2N$. For any $k$ we have

$1+x+x^{2}...+x^{k}=\frac{x^{k+1}-1}{x-1}$

and we may factor $\frac{x^{k+1}-1}{x-1}$ into cyclotomic polynomials. Thus, a
major part of understanding odd perfect numbers comes from understanding the
integer values of cyclotomic polynomials.

######  Lemma 4.

If $a$ and $b$ are distinct odd primes and $p$ is a prime such that
$p|(a^{2}+a+1)$ and $p|(b^{2}+b+1)$. If $a\equiv b\equiv 2$ (mod 3), then
$p\leq\frac{a+b+1}{5}$. If $a\equiv b\equiv 1$ (mod 3) then
$p\leq\frac{a+b+1}{3}$.

This is Lemma 1 from [23] We will also need the following result, which is
Lemma 3 in Ochem and Rao:

######  Lemma 5.

Let $p$, $q$ and $r$ be positive integers. If $p^{2}+p+1=r$ and $q^{2}+q+1=3r$
then $p$ is not an odd prime.

######  Lemma 6.

If $a$,$b$,and $c$ are odd primes and $a^{2}+a+1=3(b^{2}+b+1)(c^{2}+c+1)$,
then at least one of $b^{2}+b+1$ and $c^{2}+c+1$ is composite.

###### Proof.

Assume that $a^{2}+a+1=3(b^{2}+b+1)(c^{2}+c+1)$, and assume that $b^{2}+b+1$
and $c^{2}+c+1$ are primes. We must have either $a\equiv b$ (mod
($b^{2}+b+1$)) or $a\equiv b^{2}$ (mod ($b^{2}+b+1$)). We claim that we can
rule out the smallest few such numbers of either of these forms. We have
$a>b$, and since $a$ is prime, we cannot have $a=b^{2}$. Our next possible
value for $a$ is $b^{2}+b+1+b=(b+1)^{2}$ which is even, as is the value after
that, $b^{2}+b+1+b^{2}$.

Thus we have $a\geq 2(b^{2}+b+1)+b>2(b^{2}+b+1)$. By the same logic,
$a>2(c^{2}+c+1)$. But then we have

$3(b^{2}+b+1)(c^{2}+c+1)=a^{2}+a+1>a^{2}>4(b^{2}+b+1)(c^{2}+c+1),$

and this is a contradiction. ∎

######  Lemma 7.

If $x$ is a positive integer then the only possible common prime divisor of
$x^{2}-x+1$ and $x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3$ is 31. .

###### Proof.

Set $A=x^{2}-x+1$ and $B=x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3$. If $p|(A,B)$
then $p$ divides any linear combination of them. In particular,

$p|(5x^{5}+21x^{4}+44x^{3}+58x^{2}+55x+34)A-(5x+1)B=31.$

So $p=31$. ∎

######  Lemma 8.

The only non-negative integer solutions to the equation

$x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3=a^{2}+a+1$

$(x,a)=(0,1)$ and $(x,a)=(1,5)$.

###### Proof.

We can verify that by direct computation that these two solutions are the only
solutions where $0\leq 26\leq x$, so we may assume that $x\geq 27$. Some
algebra shows that we may write $a$ in terms of $x$ as

$a=x^{3}+\frac{3}{2}x^{2}+\frac{11}{8}x+7/16+t$ (8)

where

$t=\frac{588x^{2}+876x+351}{256x^{3}+384x^{2}+352x+240+128\sqrt{4x^{6}+12x^{5}+20x^{4}+24x^{3}+28x^{2}+24x+9}}.$

For any integer $x$, $x^{3}+\frac{3}{2}x^{2}+\frac{11}{8}x$ is a rational
number whose denominator divides $8$, and the next term in 8 is $7/16$, so the
only way that $a$ can be an integer is if $t$ is a fraction whose denominator
is $16$. However, we have that

$0<t<\frac{876x^{2}}{256x^{3}+384x^{2}}=\frac{219}{128x+96}<\frac{1}{16}.$

Here the last inequality on the right is due to $x\geq 27$. Since $t$ is
strictly between $0$ and $1/16$ it cannot be a fraction with denominator 16,
and so there are no more solutions. ∎

We then have also

######  Lemma 9.

The only positive integer solution to

$x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3=a^{2}-a+1$

is $(x,a)=(1,6)$.

###### Proof.

This follows from Lemma 8 and noting that this equation is identical to the
equation from that lemma but with $a-1$ substituted for $a$. ∎

######  Lemma 10.

Let $x$ be a prime with $\sigma(x^{4})=x^{4}+x^{3}+x^{2}+x+1$ prime, and
suppose that $a=\sigma(x^{4})$ is prime. Then $\sigma(a^{2})=a^{2}+a+1$ has at
least two distinct prime factors. Furthermore, if $\sigma(a^{2})=bc$ for two
distinct primes $b$ and $c$ then either $11|\sigma(b^{4})$ or
$11|\sigma(c^{4})$.

###### Proof.

Assume that $x$ is prime, and assume further that $a=\sigma(x^{4})$ is prime.
Then $a=x^{4}+x^{3}+x^{2}+x+1$ and we have $\sigma(a^{2})=a^{2}+a+1$. A
straightforward calculation gives us

$\begin{split}\sigma(a^{2})&=x^{8}+2x^{7}+3x^{6}+4x^{5}+6x^{4}+5x^{3}+4x^{2}+3x+3\\\
&=(x^{2}-x+1)(x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3).\end{split}$ (9)

By lemma 7 we have that $\sigma(a^{2})$ must have at least two distinct prime
factors, unless both $(x^{2}-x+1)$ and
$x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3$ are powers of 31. But they cannot
both be a power of 31. To see why note, that this would make $(x^{2}-x+1)^{3}$
also a power of 31, and we would have a contradiction due to the fact that as
long as $x>2$ we have

$(x^{2}-x+1)^{3}<x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3<31(x^{2}-x+1)^{3}.$

To prove the last part of this Lemma, we now note that if $\sigma(a^{2})=bc$
for two primes $b$ and $c$ we must have that one of the primes is $x^{2}-x+1$
and the other prime is $x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3$. Without loss
of generality, let us assume that $b=x^{2}-x+1$ and that
$c=x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3$. It is easy to check that for any
modulus we have that either $b^{4}+b^{3}+b^{2}+b+1$ or $c^{4}+c^{3}+c^{2}+c+1$
is $0$ (mod 11). ∎

######  Lemma 11.

There are no odd primes primes $x$, $a$, $b$, $c$, $d$ satisfying the
conditions:

  1. 1.

$a=\sigma(x^{4})$

  2. 2.

$\sigma(a^{2})=\sigma(b^{2})\sigma(c^{2})\sigma(d^{2})$

  3. 3.

$\sigma(b^{2})$, $\sigma(c^{2})$, and $\sigma(d^{2})$ are all prime.

###### Proof.

Assume that we have such a solution. Then we the same logic as in the proof of
Lemma 10

$\begin{split}\sigma(a^{2})&=x^{8}+2x^{7}+3x^{6}+4x^{5}+6x^{4}+5x^{3}+4x^{2}+3x+3\\\
&=(x^{2}-x+1)(x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3).\end{split}$ (10)

We have that $\sigma(b^{2})=b^{2}+b+1$, $\sigma(c^{2})=c^{2}+c+1$, and
$\sigma(d^{2})=d^{2}+d+1$. Since all three of these quantities are prime we
must have one of them equal to either $x^{2}-x+1$ or
$x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3$. Without loss of generality, we will
assume that this is $b^{2}+b+1$. We have two cases: either
$b^{2}+b+1=x^{6}+3x^{5}+5x^{4}+6x^{3}+7x^{2}+6x+3$ or $b^{2}+b+1=x^{2}-x+1$.
The first case is ruled out by Lemma 8 so we must have $b^{2}+b+1=x^{2}-x+1$.
Note that $x^{2}-x+1=(x-1)^{2}+(x-1)+1$ Thus, $b^{2}+b+1=(x-1)^{2}+(x-1)+1$
and so $b=x-1$ (since $t^{2}+t+1$ is a strictly increasing function for
$t>1/2$. But $b=x-1$ is impossible since $b$ and $x$ are both odd primes. ∎

######  Lemma 12.

We cannot have integers $a,b,c,d$ with $a\equiv b\equiv c\equiv d\equiv 1$
(mod 5) and also satisfying $a^{2}+a+1=(b^{2}+b+1)(c^{2}+c+1)(d^{2}+d+1)$.

###### Proof.

This just follows from observing that the left side of the equation is $3$
(mod 5) and the right side is $2$ (mod 5). ∎

Let $(x,a,b,c)$ be a quadruplet of odd primes all greater than 3. We say that
they form a _triple threat_ if they satisfy two conditions:

  1. 1.

We have the relationship

$x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1).$

  2. 2.

$a^{2}+a+1$, $b^{2}+b+1$, and $c^{2}+c+1$ are all prime.

If we could show that there are no triple threats in general, then we could
substantially improve our bounds in this paper for both the $3|N$ case and the
$3\not|N$ case. However, we are presently unable to do that, and so we must
satisfy ourselves with instead proving substantial enough restrictions on what
triple threats can look like. We will in particular prove that we cannot have
$x\equiv 1$ (mod 5) while also having one of $a$, $b$ or $c$ also $1$ (mod 5).
Assume that $(x,a,b,c)$ is a triple threat, and that $x\equiv a\equiv 1$ (mod
5). Then without loss of generality one must have

$(c,d)\in\\{(1,2),(2,3),(4,4)(\mathrm{mod\,}5)\\}.$ (11)

We will rule out each of these three options separately. We do by proving
various statements about the equation $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$ with
$x$, $a$, $b$, $a^{2}+a+1$, $b^{2}+b+1$ and $p$ all prime. This method of
approach has advantages. First, while triple threats do not seem to exist,
solutions of this equation do exist. Thus we will at least be proving
statements about actual, mathematical objects. Second, this equation appears
to be of natural interest for extending results beyond this paper, as will be
discussed later.

######  Lemma 13.

Assume that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$ with $x$, $a^{2}+a+1$,
$b^{2}+b+1$,$a$ and $b$ and $p$ all primes greater than 3. Assume also that
$a\leq b$. Then we have one of four possibilities:

  1. 1.

We have $a^{2}+a+1|x-a$, $b^{2}+b+1|x-b$,
$x+a+1|p(a+b+1)+\frac{x-b}{b^{2}+b+1}$, and
$x+b+1|p(a+b+1)+\frac{x-a}{a^{2}+a+1}$.

  2. 2.

We have $a^{2}+a+1|x-a$, $b^{2}+b+1|x+b+1$, and
$x+a+1|p(b-a)-\frac{x+b+1}{b^{2}+b+1}$.

  3. 3.

We have $a^{2}+a+1|x+a+1$ and $b^{2}+b+1|x-b$, and
$x+b+1|p(b-a)+\frac{x+a+1}{a^{2}+a+1}$.

  4. 4.

We have $a^{2}+a+1|x+a+1$ and $b^{2}+b+1|x+b+1$ and
$x-a|p(a+b+1)-\frac{x+b+1}{b^{2}+b+1}>0$ and
$x-b|p(a+b+1)-\frac{x+a+1}{a^{2}+a+1}$

In all four cases, the quantities on the right hand sides of the divisibility
relations are positive.

###### Proof.

Assume that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$ with $x$, $a^{2}+a+1$,
$b^{2}+b+1$, $a$, $b$ and $p$ all primes greater than 3. Then

$(x-a)(x+a+1)=x^{2}+x+1-(a^{2}+a+1)=(a^{2}+a+1)(p(b^{2}+b+1)-1)$

and

$(x-b)(x+b+1)=x^{2}+x+1-(b^{2}+b+1)=(b^{2}+b+1)(p(a^{2}+a+1)-1).$

Since $a^{2}+a+1$ and $b^{2}+b+1$ are prime we have four situations:

  1. 1.

$a^{2}+a+1|x-a$ and $b^{2}+b+1|x-b$.

  2. 2.

$a^{2}+a+1|x-a$ and $b^{2}+b+1|x+b+1$.

  3. 3.

$a^{2}+a+1|x+a+1$ and $b^{2}+b+1|x-b$.

  4. 4.

$a^{2}+a+1|x+a+1$ and $b^{2}+b+1|x+b+1$.

Note that it is easy to check that we have that $(x-a,x+a+1)=(x-b,x+b+1)=1$.
Consider each of these four situations: Case I: We have $a^{2}+a+1|x-a$ and
$b^{2}+b+1|x-b$. Set $k_{a}=\frac{x-a}{a^{2}+a+1}$ and
$k_{b}=\frac{x-b}{b^{2}+b+1}.$ Note that $x+a+1|p(b^{2}+b+1)-1$ and
$x+b+1|p(a^{2}+a+1)-1$. Note that

$x+a+1=k_{a}(a^{2}+a+1)+2a+1=k_{b}(b^{2}+b+1)+a+b+1$

and

$x+b+1=k_{a}(a^{2}+a+1)+a+b+1=k_{b}(b^{2}+b+1)+2b+1.$

We have then

$x+a+1|p(x+a+1)-k_{b}(p(b^{2}+b+1)-1)=p(k_{b}(b_{2}+b+1)+a+b+1)-k_{b}(p(b^{2}+b+1)-1).$

We then note that

$p(k_{b}(b_{2}+b+1)+a+b+1)-k_{b}(p(b^{2}+b+1)-1)=p(a+b+1)+k_{b}=p(a+b+1)+\frac{x-b}{b^{2}+b+1}.$

The other relation then follows by symmetry.
Case II: We have $a^{2}+a+1|x-a$ and $b^{2}+b+1|x+b+1$. Set

$k_{a}=\frac{x-a}{a^{2}+a+1}$

and

$k_{b}=\frac{x+b+1}{b^{2}+b+1}.$

We have then that $x+a+1|p(b^{2}+b+1)-1$. We have

$x+a+1=k_{a}(a^{2}+a+1)+2a+1=k_{b}(b^{2}+b+1)+a-b.$
$x+a+1|k_{b}(p(b^{2}+b+1)-1)-p(k_{b}(b^{2}+b+1+1)-b+a)=-(p(b-a)-k_{b}).$

We have then $x+a+1|p(b-a)-\frac{x+b+1}{b^{2}+b+1}$. We need to show that
$p(b-a)-\frac{x+b+1}{b^{2}+b+1}>0$. Assume that $p(b-a)-k_{b}\leq 0.$ Note in
this case we have $b\neq a$, and so $p(b-a)\leq\frac{x+b+1}{b^{2}+b+1}$. Since
$a\equiv b\equiv 2$ (mod 6), we have that $b-a\geq 6$. Thus,
$6p\leq\frac{x+b+1}{b^{2}+b+1}$, and we get that $2pb^{2}<x$. Then

$2pa^{2}<2pb^{2}<x.$

We then have

$4p^{2}a^{2}b^{2}<x^{2}<x^{2}+x+1=pa^{2}b^{2},$

which is a contradiction. Thus, $p(b-a)-k_{b}$ is positive. Case III: We have
that

$a^{2}+a+1|x+a+1$

and

$b^{2}+b+1|x-b.$

We set $k_{a}=\frac{x+a+1}{a^{2}+a+1}$ and $k_{b}=\frac{x-b}{b^{2}+b+1}.$ We
have then from logic identical to that in Case II that $x+b+1|p(a-b)-k_{a}$.
The right hand side is negative, so $x+b+1|p(b-a)+k_{a}$ is positive.

Case IV: We have $a^{2}+a+1|x+a+1$ and $b^{2}+b+1|x+b+1$.
We have then $x-a|p(x-a)-k_{b}(p(b^{2}+b+1)-1$ where
$k_{b}=\frac{x+b+1}{b^{2}+b+1}$. From $x-a=k_{b}(b^{2}+b+1)-a-b-1$ we get that
$x-a|p(a+b+1)-k_{b}$. The other case follows from similar reasoning.
Positivity follows from an argument similar to the case in Case II.

∎

######  Lemma 14.

Assume that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$, with $x$, $a^{2}+a+1$,
$b^{2}+b+1$,$a$ and $b$ and $p$ all primes greater than 3 and $a\leq b$. .
Then $2b<p$.

###### Proof.

Note that we must have that $x\equiv a\equiv b\equiv 2$ (mod 3) We split up
into each of the four cases from Lemma 1

Case I: We have that $x+a+1|p(a+b+1)+k_{b}$ where
$k_{b}=\frac{x-b}{b^{2}+b+1}$. Thus $x+a+1\leq p(a+b+1)+k_{b}$. This gives us
that

$k_{b}(b^{2}+b+1)+a+b+1\leq p(a+b+1)+k_{b}.$

This is the same as

$\frac{k_{b}b^{2}}{a+b+1}+\frac{k_{b}b}{a+b+1}+1\leq p.$

Note that $a+b+1\leq 2b$ and that congruence arguments give us that $k_{b}\geq
6$. We have in this case that $a\neq b$, and so $a+b+1<2b+1<3$. We get that

$2b+2<\frac{k_{b}b^{2}}{a+b+1}+\frac{k_{b}b}{a+b+1}+1\leq p.$

Case II:

We have $x+a+1|p(b-a)-k_{b}$ with $k_{b}=\frac{x+b+1}{b^{2}+b+1}$. Note that
$p(b-a)-k_{b}$ is positive. We have that $p(b-a)-k_{b}\equiv 1$ (mod 6), and
$x+a+1\equiv 5$ (mod 6). Therefore $h(x+a+1)=p(b-a)-k_{b}$ for some $h\geq 5$.
Thus we have

$5(x+a+1)\leq p(b-a)-k_{b}.$

Since
$k_{b}=\frac{x+b+1}{b^{2}+b+1}=\frac{x}{b^{2}+b+1}+\frac{b+1}{b^{2}+b+1}$ and
$b>3$, we have that $k_{b}<x/12$. Thus

$5x\leq p(b-a)-x/7.$

Then $\frac{36}{5}x\leq p(b-a)$. Since $b^{2}+b+1\leq x$, the desired
inequality follows. Case III: Case III is very similar to case II. We have
that $x+b+1|p(b-a)+k_{a}$ where $k_{a}=\frac{x+a+1}{a^{2}+a+1},$ We have that
$x+b+1\leq p(b-a)+k_{a}$. We have $k_{a}\leq x/13$, and so we have that

$(12/13)x\leq p(b-a)-b-1.$

From $b^{2}+b+1|x-b,$ we get that $b^{2}+b+1\leq\frac{x}{6}$ which gives us
the desired inequality.
Case IV: We have then $x-a|p(x-a)-k_{b}(p(b^{2}+b+1)-1$ where
$k_{b}=\frac{x+b+1}{b^{2}+b+1}$. From $x-a=k_{b}(b^{2}+b+1)-a-b-1$ we get that
$x-a|p(a+b+1)-k_{b}$. Thus $k_{b}(b^{2}+b+1)-a-b-1\leq p(a+b+1)-k_{b}$ and so
we get that from $k_{b}\geq 5$ and $k_{b}b^{2}<p(a+b+1)$ that $2b<p$. ∎

We get as an immediate corollary:

######  Corollary 15.

Assume that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$. Assume that $a^{2}+a+1$,
$b^{2}+b+1$, and $p>3$ are all prime. Then $p>x^{2/5}$.

###### Proof.

Assume that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$, with $a^{2}+a+1$,
$b^{2}+b+1$, and $p$ prime. As usual, assume that $a\leq b$. Then we have that
$x^{2}+x+1=p(a^{2}+a+1)(b^{2}+b+1)<p(\frac{p^{2}}{4}+\frac{p}{2}+1)(\frac{p^{2}}{4}+\frac{p}{2}+1)\leq\frac{3}{4}p^{5}$
from which the desired inequality follows. ∎

######  Lemma 16.

If $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$, with $a^{2}+a+1$, $b^{2}+b+1$, $a$,
$x$ and $p$ are all primes prime greater than 3, and $b\geq a\geq 5$, then
$a^{2}+a+1<\frac{49}{4}p$.

###### Proof.

This proof requires breaking down the cases above in further detail. In Case
I, $x+b+1\leq p(a+b+1)+k_{a}$. Note that $k_{a}\leq\frac{x}{7}$ which yields

$\frac{6}{7}x\leq p(a+b+1)-b-1.$

So

$x\leq\frac{7}{6}(p(a+b+1)-b-1)\leq\frac{7}{6}p(3b)-1=\frac{7}{2}pb-1.$

We have then that

$p(a^{2}+a+1)b^{2}<p(a^{2}+a+1)(b^{2}+b+1)=x^{2}+x+1\leq(\frac{7}{2}(pb-1))^{2}+\frac{7}{2}(pb)-1+1<\frac{49}{4}p^{2}b^{2}$

which implies that $a^{2}+a+1<\frac{49}{4}p.$
The other cases are similar. ∎

######  Lemma 17.

If $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$, with $a^{2}+a+1$, $b^{2}+b+1$, and
$p>3$ prime and $b\geq a\geq 3$, then $a^{2}+a+1<(25/2)^{1/3}x^{2/3}\leq
3x^{2/3}$.

###### Proof.

We have

$x^{2}+x+1=p(a^{2}+a+1)(b^{2}+b+1)\geq\frac{4}{49}(a^{2}+a+1)^{3}$

and so

$(50/4)x^{2}\geq(a^{2}+a+1)^{3}$

which leads to the desired inequalities. ∎

Note that this means that we can take $k_{a}\geq\frac{x^{1/3}}{3}$ in all
cases of Lemma 14 and can get tighter versions of that result.

######  Lemma 18.

Assume that $x$ is a prime. Then $x^{2}+x+1$ is not a perfect cube.

###### Proof.

Assume that $x$ is prime and $x^{2}+x+1=k^{3}$. Then we have that

$x(x+1)=k^{3}-1=(k?1)(k^{2}+k+1).$

Since $x$ is prime, either $x|k-1$ or $x|k^{2}+k+1.$ If $x|k-1$, then $x\leq
k-1$ and $k^{3}\geq(x+1)^{3}>x^{2}+x+1=k^{3}$, which is impossible. Now
consider the possibility that $x|k^{2}+k+1.$ Since $x$ is prime, then either
$x=3$, which does not give a solution , or $x\equiv 1$ (mod 3). Thus,
$x^{2}+x+1\equiv 3$ (mod 9), but no perfect cube is congruent to $3$ (mod 9)
and so we again reach a contradiction. ∎

######  Lemma 19.

Assume that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$, with $a^{2}+a+1$,
$b^{2}+b+1$, $a$, and $b$ all primes greater than $3$. Assume that $b\geq$.
Further assume that $x\equiv a\equiv b\equiv 1$ (mod 5). Then
$\frac{11}{3}b<p$.

###### Proof.

This follows the same sort of logic as Lemma 14. In Case I we get that

$\frac{k_{b}b^{2}}{a+b+1}+\frac{k_{b}b}{a+b+1}+1\leq p.$

Note that congruence arguments give us that $k_{b}\geq 6$ and thus $a+b+1\leq
2b+1<3b$. We have by congruence arguments that $k_{b}\geq 30$, and so $10b<p$.
In Case II, we have as before $\frac{34}{35}x\leq p(b-a)$ and
$x+b+1=k_{b}(b^{2}+b+1)$. Note that $k_{b}$ is odd. We have that $x+b+1\equiv
2$ (mod 3), and $b^{2}+b+1\equiv 1$ (mod 3) so $k_{b}\equiv 2$ (mod 3).
Similarly, we have that $k_{b}\equiv 1$ (mod 5). So $k_{b}\geq 11$. We have
then

$b^{2}+b+1\leq\frac{x+b+1}{11}.$

This is the same as

$11b^{2}+10b+10\leq x,$

and so

$11b^{2}+10b+10\leq\frac{35}{34}p(b-a),$

which is stronger than the desired inequality.
In Case III, we have $(12/13)x\leq p(b-a)-b-1.$ and
$k_{b}=\frac{x-b}{b^{2}+b+1}.$ We have then $30|k_{b}$, and so

$b^{2}+b+1\leq\frac{x-b}{30}$

and so $30b^{2}+31b+30\leq x$, and combining as before yields an inequality
stronger than the one in equation.
In Case IV, we that $k_{b}=\frac{x+b+1}{b^{2}+b+1}$ as in Case II. As in Case
II, we get that $k_{b}$ is odd, $k_{b}\equiv 2$ (mod 3) and $k_{b}\equiv 1$
(mod 5). So $k_{b}\geq 11$. In case IV we had that $k_{b}b^{2}<p(a+b+1)$ and
since $a+b+1<3b$ this becomes

$11b^{2}<3pb$

which implies that

$\frac{11}{3}b<p.$

∎

######  Lemma 20.

Suppose that $x,a,b,p$ are all primes greater than 3 where

$x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p.$

Suppose also that $a^{2}+a+1$, $b^{2}+b+1$ are prime, and that $a\leq b$. Then
in Case II and Case III we have $a^{2}+a+1<\frac{p}{16}$.

###### Proof.

Assume we are in Case II. So we have $x+a+1|p(b-a)-\frac{x+b+1}{b^{2}+b+1}$.
We note that $x+a+1\equiv 2$ (mod 3) and $p(b-a)-\frac{x+b+1}{b^{2}+b+1}\equiv
1$ (mod 3). Note also that both quantities are odd. We thus have
$m(x+a+1)=p(b-a)-\frac{x+b+1}{b^{2}+b+1}$ for some $m\geq 5$. We have then

$5(x+a+1)\leq p(b-a)-\frac{x+b+1}{b^{2}+b+1}.$

We have then $x<\frac{pb}{5}-2.$ Then

$(a^{2}+a+1)(b^{2}+b+1)p=x^{2}+x+1=(\frac{pb}{5}-2)^{2}+\frac{pb}{5}-2+1<\frac{p^{2}b^{2}}{25}.$

So we have

$(a^{2}+a+1)(b^{2}+b+1)p<\frac{p^{2}b^{2}}{25}$

and so we have that $a^{2}+a+1<\frac{p}{25}$ which implies the desired result.
Now for Case III: We have that $x+b+1|p(b-a)+\frac{x+a+1}{a^{2}+a+1}.$ By
similar logic as above we have that $m(x+b+1)=p(b-a)+\frac{x+a+1}{a^{2}+a+1}$
where $m\equiv 5$ (mod 6). We have $5(x+b+1)\leq
p(b-a)+\frac{x+a+1}{a^{2}+a+1}$ and so

$x<\frac{pb}{4}-2$

Then

$(a^{2}+a+1)(b^{2}+b+1)p=x^{2}+x+1=(\frac{pb}{4}-2)^{2}+\frac{pb}{4}-2+1<\frac{p^{2}b^{2}}{16},$

from which the result follows. ∎

######  Lemma 21.

Suppose that $x,a,b$ and $p$ are all primes greater than 3 where

$x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p.$

and $a^{2}+a+1$, $b^{2}+b+1$ are prime, and that $a\leq b$. Assume further
that we have Case IV and that we have $x-a=p(a+b+1)-k_{b}$ and
$x-b=p(a+b+1)-k_{a}$. Then we must have $a=b$ and $a^{2}+a+2|7(p+1).$

###### Proof.

Assume as given. We have then

$(x-b)(a^{2}+a+1)+x+a+1=p(a+b+1)(a^{2}+a+1)$ (12)

and

$(x-a)(b^{2}+b+1)+x+b+1=p(a+b+1)(b^{2}+b+1).$ (13)

Assume for now now that $a\neq b$. from Equations 12 and 13 that

$\frac{(x-b)(a^{2}+a+1)+x+a+1}{a^{2}+a+1}=p(a+b+1)=\frac{(x-a)(b^{2}+b+1)+x+b+1}{b^{2}+b+1}.$
(14)

We can solve the above for $x$ to get that

$x=\frac{(b-a)(a^{2}+a+1)(b^{2}+b+1)-(b+1)(a^{2}+a+1)+(a+1)(b^{2}+b+1)}{b^{2}+b-a^{2}-a}.$
(15)

Now, it turns out the top and bottom both have a factor of $b-a$ and this is a
meaningful solution for $x$ because we have assumed that $b\neq a$.
Simplifying we get that

$x=\frac{a^{2}b^{2}+a^{2}b+a^{2}+ab^{2}+2ab+2a+b^{2}+2b+1}{a+b+1},$

and this forces $x$ to be even which is a contradiction since $x$ is an odd
prime. We may thus assume that $a=b$.

Thus, both Equation 12 and Equation 13 become the same thing:

$(x-a)(a^{2}+a+1)+x+a+1=p(2a+1)(a^{2}+a+1).$ (16)

We may rearrange Equation 16 to obtain

$x(a^{2}+a+2)=(p(2a+1)+a)(a^{2}+a+1)-a-1.$ (17)
$a^{2}+a+2|(p(2a+1)+a)(a^{2}+a+1)-a-1.$ (18)

. We also trivially have

$a^{2}+a+2|(p(2a+1)+a)(a^{2}+a+2).$ (19)

We may then conclude that $a^{2}+a+2$ divides the difference of 18 and 19. So
$a^{2}+a+2$ divides $p(2a+1)+2a+1=(p+1)(2a+1)$. It is easy to check that
$gcd(a^{2}+a+2,2a+1)|7$. Thus, we have

$a^{2}+a+2|7(p+1).$ (20)

which was what was claimed. ∎

######  Lemma 22.

Suppose that $x,a,b$ and $p$ are all primes greater than 3 where

$x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p.$

Suppose further that $x\equiv a\equiv b\equiv 1$ (mod 5), and $a^{2}+a+1$,
$b^{2}+b+1$ are prime, and that $a\leq b$, and $p\geq 47$. Then we have
$a^{2}+a+1<p$.

###### Proof.

We will as before break into 4 cases in the same way as before. Cases II and
III are handled by Lemma 20 In this situation, almost all the serious work
will be in Case IV. In Case I: We have as before that $x+b+1|p(a+b+1)+k_{a}$
where $k_{a}=\frac{x-a}{a^{2}+a+1}$. Set $m(x+b+1)=p(a+b+1)+k_{a}$. We have
that $x+b+1\equiv 3$ (mod 5) and that $p(a+b+1)+k_{a}\equiv 1$ (mod 5). Thus,
we have $m\equiv 2$ (mod 5). Similarly, we have that $m\equiv 1$ (mod 3). So
$m\geq 7$ and so $7(x+b+1)\leq p(a+b+1)+k_{a}$. Since $a^{2}+a+1>7$, we have
that $k_{a}\leq\frac{x}{7}$, so

$\frac{48}{7}x\leq p(a+b+1)-7b-7$

which yields

$x\leq\frac{7}{48}p(a+b+1)-\frac{49}{48}b-\frac{49}{48}.$

We have that $a+b+1<3b$ and so

$x\leq\frac{7}{16}pb-\frac{49}{48}-\frac{49}{48}\leq\frac{7}{16}pb-2.$

We have then

$(a^{2}+a+1)(b^{2}+b+1)p=x^{2}+x+1\leq(\frac{7}{16}pb-2)^{2}+\frac{7}{16}pb-2+1<\left(\frac{7}{16}\right)^{2}(pb^{2}).$

From the above we get that $a^{2}+a+1\leq\left(\frac{7}{16}\right)^{2}p<p.$
Now for Case IV. We have $(x-a)m_{a}=p(a+b+1)-k_{b}$ and
$(x-b)m_{b}=p(a+b+1)-k_{a}$ for some $m_{a}$ and $m_{b}$. In the first case,
assume that $m_{a}=m_{b}=1$.

Then by Lemma 21 we have that $k(a^{2}+a+2)=7(p+1)$ for some $k$. Note that
$a^{2}+a+2\equiv 4$ (mod 5). We also have $7(p+1)\equiv 1$ (mod 5). Thus, we
have $k\equiv 4$ (mod 5). Similarly, we have that $k\equiv 1$ (mod 3). So
$k\equiv 4$ (mod 15). Consider the possibility of $k=4$. If that is the case
then we have $4(a^{2}+a+2)=7(p+1)$, and this is the same as

$7p=4a^{2}+4a+1=(2a+1)^{2}.$

But then we must have $p=7$ and $a=3$, which is ruled out by our initial
assumption that $a>3$. Thus, since $k\equiv 4$ (mod 15), we get that $k\geq
19$. In that case we have $19(a^{2}+a+1)\leq 7(p+1)$, from which it follows
that $a^{2}+a+1<p.$ Next we will consider $m_{a}=2$ and $m_{b}=1$. We need to
consider then the equation

$2(x-a)=p(a+b+1)-k_{b}.$

Since $m_{b}\neq m_{a}$ we have that $a\neq b$, and thus $a\leq b-30$ (since
$a$ and $b$ agree (mod 30)). So we have that

$x=\frac{p(a+b+1)}{2}-\frac{k_{b}}{2}+a\leq\frac{p(2b-29)}{2}-\frac{k_{a}}{2}+a.$

We note that Lemma 14 implies that $a<p$, and so we have that

$x\leq pb-\frac{29p}{2}+p<pb-2,$

from which the same logic as used in Case I holds. The case of $m_{b}=2$ and
$m_{a}=1$ is nearly identical, as is the case of either $m_{a}$ or $m_{b}$
being 2 but with $a\neq b$. We will then next consider the case
$m_{a}=m_{b}=2$, and $a=b$. We have

$2(x-a)=p(2a+1)-\frac{x+a+1}{a^{2}+a+1}.$

Solving for $x$ we obtain:

$(2a^{2}+2a+3)x=(p(2a+1)+2a)(a^{2}+a+1)-a-1.$

Thus, we have $2a^{2}+2a+3|(p(2a+1)+2a)(a^{2}+a+1)-a-1$, and we get that

$2a^{2}+2a+3|(p(2a+1)+2a)(2a^{2}+2a+2)-2a-2.$ (21)

Trivially,

$2a^{2}+2a+3|(p(2a+1)+2a)(2a^{2}+2a+3).$

Taking the difference between the quantities on the right-hand sides of
Equation 21 and Equation 1, we get that

$2a^{2}+2a+3|p(2a+1)+4a+2.$

We note that both quantities in the above agree mod 10. So there is some
$m\equiv 1$ (mod 10) such that

$m(2a^{2}+2a+3)=p(2a+1)+4a+2.$

We claim that $m\neq 1$. To see this, assume that $m=1$, so we have

$2a^{2}+2a+3=p(2a+1)+4a+2,$

which is the same as

$2a^{2}-2a+1=p(2a+1).$

Thus we have $2a+1|2a^{2}-2a+1$. It is easy to check that the only positive
integer which satisfies this is $a=2$. This is impossible since $a$ is an odd
prime. By the above remarks we must have either $m_{a}\geq 3$ or $m_{b}\geq
3$. We will consider $m_{a}\geq 3$ (the case for $m_{a}\geq 3$ is essentially
identical). We have

$3(x-a)\leq p(a+b+1)-k_{b}.$

We note that $a+b+1\leq 2b+1$, and thus

$x\leq\frac{p(2b+1)}{3}-k_{b}+a.$

We note that $p\geq 47$, together with Lemma 16 gives us that $a<1/3p$. We
again obtain that

$x\leq pb-2,$

and again draw the same conclusion. ∎

######  Lemma 23.

Suppose that we have odd primes $x,a,b,p$ all primes greater than 3 where

$x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p.$

Suppose further that $x\equiv a\equiv 1$ (mod 5), $b\equiv 2$ (mod 5) and
$a^{2}+a+1$, $b^{2}+b+1$ are prime, and that $a\leq b$, and $p\geq 47$. Then
we have $a^{2}+a+1<p$.

###### Proof.

As usual, we will have four cases to check, and as in the last lemma most of
the difficulty will be in Case IV. In this lemma as in the last one we will be
able to rely on the previous results concerning this situation, but note that
due to the different congruence assumption we cannot here make use of Lemma 22
Note that in this lemma we now have $p\equiv 3$ (mod 5), rather than in the
previous lemma where we had $p\equiv 2$ (mod 5). Case I: We have as before
that $x+b+1|p(a+b+1)+\frac{x-a}{a^{2}+a+1}$. Set
$m(x+b+1)=p(a+b+1)+\frac{x-a}{a^{2}+a+1}$. We have that
$p(a+b+1)+\frac{x-a}{a^{2}+a+1}\equiv 2$ (mod 5) and $x+b+1\equiv 4$ (mod 5).
Thus $m\equiv 3$ (mod 5). Note that $m$ is odd, and also that $(3,m)=1$, and
so $m\geq 13$. We have then

$13(x+b+1)\leq p(a+b+1)+\frac{x-a}{a^{2}+a+1}.$

The same logic as in the previous lemma then applies. Lemma 20 handles Case II
and Case III.

Case IV: We have $a^{2}+a+1|x+a+1$ and $b^{2}+b+1|x+b+1$. Note that unlike in
Lemma 22 we have immediately that $a\neq b$ because they disagree mod $5$. We
have $x-a|p(a+b+1)-k_{b}$ and $x-b|p(a+b+1)-k_{a}$ where
$k_{b}=\frac{x+b+1}{b^{2}+b+1}$, and $k_{a}=\frac{x+a+1}{a^{2}+a+1}.$ We have
$(x-a)m_{a}=p(a+b+1)-k_{b}$ and $(x-b)m_{b}=p(a+b+1)-k_{a}$ for some positive
integers $m_{a}$ and $m_{b}$. We will consider various possible options for
the pair $(m_{a},m_{b})$. The pair $(1,1)$ is already ruled out since then
Lemma 21 would force $a=b$.

Since we have that $a\neq b$, and thus $a\leq b-6$ (since $a<b$ and $a\equiv
b+1$ (mod 30)). So we have that

$x=\frac{p(a+b+1)}{2}-\frac{k_{b}}{2}+a\leq\frac{p(2b-6)}{2}-\frac{k_{a}}{2}+a.$

Note that Lemma 14 implies that $a<p$, and so we have that

$x\leq pb-3p+p<pb-2,$

from which the same logic as used in Case I holds. For the remaining possible
options for $(m_{a},m_{b})$ we follow logic essentially identically to those
in remaining parts of the Case IV of the proof of Lemma 22. ∎

Using nearly identical logic to the above lemma we have:

######  Lemma 24.

Suppose that we have odd primes $x,a,b,p$ all primes greater than 3 where

$x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p.$

Suppose further that $a\equiv 2$ (mod 5), $x\equiv b\equiv 1$ (mod 5). Assume
that $a^{2}+a+1$ and $b^{2}+b+1$ are prime. Finally, assume that $a\leq b$,
and $p\geq 47$. Then $a^{2}+a+1<p$.

######  Lemma 25.

There are no solutions to the equation

$x^{2}+x+1=(p^{2}+p+1)(q^{2}+q+1)(r^{2}+r+1)$

with $x$, $p$, $q$, $r$, $p^{2}+p+1$, $q^{2}+q+1$, $r^{2}+r+1$ all prime and
with $x\equiv p\equiv q\equiv 1$ (mod 5). That is, there does not exist any
triple threat $(x,p,q,r)$ where $x\equiv p\equiv q\equiv 1$ (mod 5).

###### Proof.

Assume we have a solution. We note that we must also have $r\equiv 2$ (mod 5).
We must have $\min(p^{2}+p+1,q^{2}+q+1,q^{2}+q+1)>47$. If any were not, we
could use Corollary 15 to conclude that we have $47>x^{2/5}$, this gives us
only a finite set of $x$ to check and we can easily verify that none of them
are solutions. We may without loss of generality also assume that $p\leq q$.
We conclude by Lemma 22 that $p^{2}+p+1<r^{2}+r+1$, and so $p<r$. We may apply
Lemma 23 to get that $p<q$. We have two cases to consider. Either $p<r<q$ or
$p<q<r$. If $p<r<q$, then Lemma 23 gives us that $r^{2}+r+1<p^{2}+p+1$ and
hence $r<p$ which is impossible. So we may assume that $p<q<r$, but then by
Lemma 24 we have that $q^{2}+q+1<p^{2}+p+1$ and hence $q<p$ which is
impossible. So all cases have lead to a contradiction. ∎

The basic thrust of the next set of results is very similar.

######  Lemma 26.

Suppose that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$ where
$x,a,b,a^{2}+a+1,b^{2}+b+1$ and $p$ are odd primes greater than 3. Suppose
further that $p=c^{2}+c+1$ for some $c\equiv 5$ (mod 6) . Suppose that $a\leq
b$ and that we are in Case I. Then $a^{2}+a+1<\frac{p}{4}$.

###### Proof.

Assume as given. So we have $x+a+1|p(a+b+1)+\frac{x-b}{b^{2}+b+1}$ and
$x+b+1|p(a+b+1)+\frac{x-a}{a^{2}+a+1}$. We may set
$m_{a}(x+a+1)=p(a+b+1)+\frac{x-b}{b^{2}+b+1}$ and
$m_{b}(x+b+1)=x+b+1|p(a+b+1)+\frac{x-a}{a^{2}+a+1}$ We’ll first assume that
$m_{a}=m_{b}=1$ and then handle the remaining cases. If $m_{a}=m_{b}=1$ then
we have

$x+a+1=p(a+b+1)+\frac{x-b}{b^{2}+b+1}$ (22)

and

$x+b+1=p(a+b+1)+\frac{x-a}{a^{2}+a+1}$ (23)

We’ll first assume that $a\neq b$, and arrive at a contradiction. We’ll then
handle $a=b$ (which is where we will need the assumption that $p=c^{2}+c+1$).
Assume that $a\neq b$. Then we may subtract Equation 22 from Equation 23 to
get that:

$b-a=\frac{x-a}{a^{2}+a+1}-\frac{x-b}{b^{2}+b+1}.$

This is the same as

$(a^{2}+a+1)(b^{2}+b+1)(b-a)=x(b-a)(a+b+1)+ab(b-a)+(b-a).$

Since $b\neq a$ we have $b-a\neq 0$ and so we may divide by $b-a$ to get

$(a^{2}+a+1)(b^{2}+b+1)=x(a+b+1)+ab+1.$

This is the same as
$x=\frac{a^{2}b^{2}+a^{2}b+b^{2}a+a^{2}+b^{2}+a+b}{a+b+1}$. But we must have
$a\equiv b\equiv 2$ (mod 3) and also $x\equiv 2$ (mod 3). But $a\equiv b\equiv
2$ (mod 3) forces the right hand-side of the above to be $1$ (mod 3). So we
must have $a=b.$ Since $a=b$ we have

$x+a+1=p(2a+1)+\frac{x-a}{a^{2}+a+1}.$

which can be rearranged to

$(p-1)(2a^{2}+3a^{2}+3a+1)=-a^{3}+a^{2}(x-a)+ax.$

We have that $(a,2a^{2}+3a^{2}+3a+1)=1$ and so $a|p-1$. Now since
$p=c^{2}+c+1$ this is the same as saying that $a|c(c+1)$. Either $a|c$ or
$a|c+1$. If $a|c$ then either $a=c$, or $a<c$. If $a=c$ and so $p=a^{2}+a+1$
and $x^{2}+x+1=(a^{2}+a+1)^{3}$. But this would contradict Lemma 18. If $a<c$,
then since $a|c$ and $a\equiv c\equiv 5$ (mod 6), so we would have $7a\leq c$,
from which it easily follows that $a^{2}+a+1<\frac{p}{4}$. If we have $a|c+1$,
then since $6|c+1$ and we have that $6a|c+1$ and so $6a\leq c+1$, from which
it easily follows that $a^{2}+a+1<\frac{p}{4}$ We now need to handle the case
o $m_{a}$ and $m_{b}$ are not both equal to 1. We will look at the case of
$m_{a}\neq 1$ ($m_{b}\neq 1$ is essentially identical). We have that
$m_{a}(x+a+1)=p(a+b+1)+\frac{x-b}{b^{2}+b+1}$ for some $m_{a}>1$. We note that
we cannot have $m_{a}$ even because the right-hand side of the equation is
odd. We also cannot have $m_{a}=3$ because the right hand-side is $2$ (mod 3).
We therefore have $m_{a}\geq 5$. We have then: $5(a+x+1)\leq
p(a+b+1)+\frac{x-b}{b^{2}+b+1}$ from which the desired inequality follows. ∎

We would like in Lemma 26 to remove the need for the assumption that
$p=c^{2}+c+1$ but for our results here at least that is not necessary.

######  Lemma 27.

Suppose that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$ where $x,a,b,p,a^{2}+a+1$ and
$b^{2}+b+1$ are all prime. Assume that $x\equiv a\equiv 1$ (mod 5), and that
$b\equiv 4$ (mod 5), and that $a<b$. Assume also that $p=c^{2}+c+1$ for some
$c\equiv 5$ (mod 6). Then $a^{2}+a+1<\frac{p}{4}$

###### Proof.

Note that $p\equiv 1$ (mod 5). We again split into four cases. Case I is
handled by Lemma 26. As usual, Cases II and III are handled by Lemma 20. So we
need only concern ourselves with Case IV. In Case IV we have:
$x-b|p(a+b+1)-\frac{x+a+1}{a^{2}+a+1}$. We have then for some $m_{b}$ that

$m_{b}(x-b)=p(a+b+1)-\frac{x+a+1}{a^{2}+a+1}.$

Note $x-b\equiv 2$ (mod 5), and $p(a+b+1)-\frac{x+a+1}{a^{2}+a+1}\equiv 0$
(mod 5). So $m_{b}\geq 5$ We have then that $5(x-b)\leq
p(a+b+1)-\frac{x+a+1}{a^{2}+a+1}$ from which the desired bound follows. ∎

Using nearly identical logic we have:

######  Lemma 28.

Suppose that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$ where $x,a,b,p,a^{2}+a+1$ and
$b^{2}+b+1$ are all prime. Assume that $x\equiv b\equiv 1$ (mod 5), $a\equiv
4$ (mod 5), and $a<b$. Assume also that $p=c^{2}+c+1$ for some $c\equiv 5$
(mod 6). Then $a^{2}+a+1<\frac{p}{4}$.

######  Lemma 29.

Suppose that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$ where $x,a,b,p,a^{2}+a+1$ and
$b^{2}+b+1$ are all prime. Assume that $x\equiv 1$ (mod 5), that $a\equiv
b\equiv 4$ (mod 5) and that $p\equiv 3$ (mod 5), and that $a<b$. Then
$a^{2}+a+1<\frac{p}{16}$.

###### Proof.

We have our four cases as usual with Cases II and III handled by Lemma 20. In
Case I we have We have as before that $x+b+1|p(a+b+1)+\frac{x-a}{a^{2}+a+1}$
Set $m(x+b+1)=p(a+b+1)+\frac{x-a}{a^{2}+a+1}$. We have that $x+b+1\equiv 1$
(mod 5) and $p(a+b+1)+\frac{x-a}{a^{2}+a+1}=4$ (mod 5). So $m\equiv 4$ (mod
5). Since $(6,m)=1$ we have that $m\geq 19$. We have then that $19(x+b+1)\leq
p(a+b+1)+\frac{x-a}{a^{2}+a+1}$ from which the desired inequality follows.
In Case IV we have $x-b|p(a+b+1)-\frac{x+a+1}{a^{2}+a+1}$. We set
$m(x-b)=p(a+b+1)-\frac{x+a+1}{a^{2}+a+1}$. We note that $x-b\equiv 2$ (mod 5),
and $p(a+b+1)-\frac{x+a+1}{a^{2}+a+1}\equiv 1$ (mod 5). Thus, $m\equiv 2$ (mod
5) and so $m\geq 7$. We have then that $7(x-b)\leq
p(a+b+1)-\frac{x+a+1}{a^{2}+a+1}$ from which the desired inequality follows. ∎

######  Lemma 30.

There are no solutions to the equation

$x^{2}+x+1=(p^{2}+p+1)(q^{2}+q+1)(r^{2}+r+1)$

with $x$, $p$, $q$, $r$, $p^{2}+p+1$, $q^{2}+q+1$, $r^{2}+r+1$ all prime and
with $x\equiv p\equiv 1$ (mod 5) and $q\equiv r\equiv 4$ (mod 5). That is,
there does not exist any triple threat $(x,p,q,r)$ where $x\equiv p\equiv 1$
(mod 5) and $q\equiv r\equiv 4$ (mod 5).

###### Proof.

Assume we have such a solution. Without loss of generality we may assume that
$q\leq r$. We have then from Lemma 29 that $q<p$. We thus have either $q<p<r$
or $q<r<p$ (we cannot have $p=r$ since they disagree mod 5). If we have that
$q<p<r$, then from Lemma 27 we have that $p<q$ which is a contradiction. If we
have $q<r<p$ then by Lemma 28 we have that $r<q$ which is a contradiction.
Since all possibilities lead to a contradiction, the corresponding type of
triple threat cannot exist. ∎

We now turn our attention to our final type of triple threat.

######  Lemma 31.

Suppose that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$ where
$x,a,b,a^{2}+a+1,b^{2}+b+1$ and $p$ are primes. Suppose further that $x\equiv
b\equiv 1$ (mod 5), $a\equiv 3$ (mod 5), and $p\equiv 2$ (mod 5). Suppose that
$a<b$. Then we have that $a^{2}+a+1<p$.

###### Proof.

We split into four cases as usual with cases II and III handled by lemma 20.
In Case I we have that $x+b+1|p(a+b+1)+\frac{x-b}{b^{2}+b+1}$. We have that
$m(x+b+1)=p(a+b+1)+\frac{x-b}{b^{2}+b+1}$ for some $m$. We note that
$x+b+1\equiv 3$ (mod 5) and $p(a+b+1)+\frac{x-b}{b^{2}+b+1}\equiv 1$ (mod 5).
We have then that $m\equiv 2$ (mod 5). Since $m$ is odd, we have that $m\geq
7$, and $7(x+b+1)\leq p(a+b+1)+\frac{x-b}{b^{2}+b+1}$, from which the desired
inequality follows. In Case IV we have that
$x-a|p(a+b+1)-\frac{x+b+1}{b^{2}+b+1}$. We have that for some $m$,
$m(x-a)=p(a+b+1)-\frac{x+b+1}{b^{2}+b+1}$. We have that $x-a\equiv 3$ (mod 5),
and $p(a+b+1)-\frac{x+b+1}{b^{2}+b+1}\equiv 4$ (mod 5). We have then that
$m\equiv 3$ (mod 5). We have then that $3(x-b)\leq
p(a+b+1)-\frac{x+b+1}{b^{2}+b+1}$ from which the inequality follows. ∎

Using nearly identical logic we have:

######  Lemma 32.

Suppose that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$ where
$x,a,b,a^{2}+a+1,b^{2}+b+1$ and $p$ are primes. Suppose further that $x\equiv
a\equiv 1$ (mod 5), $b\equiv 3$ (mod 5), and $p\equiv 2$ (mod 5). Suppose that
$a<b$. Then we have that $a^{2}+a+1<p$.

######  Lemma 33.

. Suppose that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$ where
$x,a,b,a^{2}+a+1,b^{2}+b+1$ and $p$ are primes. Suppose further that $x\equiv
1$ (mod 5), $a\equiv 2$ (mod 5), $b\equiv 3$ (mod 5), and $p\equiv 3$ (mod 5).
Suppose that $a<b$. Then we have that $a^{2}+a+1<p$.

###### Proof.

We again split into four cases and handle cases II and III via lemma 20. In
Case I, we have that $x+a+1|p(a+b+1)+\frac{x-b}{b^{2}+b+1}$. We have that
$x+a+1\equiv 4$ (mod 5), and $p(a+b+1)+\frac{x-b}{b^{2}+b+1}\equiv 1$ (mod 5).
The rest of the case follows as usual. For Case IV we have that
$x-a|p(a+b+1)-\frac{x+b+1}{b^{2}+b+1}$. We have that $x-a\equiv 4$ (mod 5),
and that $p(a+b+1)-\frac{x+b+1}{b^{2}+b+1}\equiv 3$ (mod 5), and the rest of
the argument follows as usual. ∎

By nearly identical logic we have:

######  Lemma 34.

. Suppose that $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)p$ where
$x,a,b,a^{2}+a+1,b^{2}+b+1$ and $p$ are primes. Suppose further that $x\equiv
1$ (mod 5), $a\equiv 3$ (mod 5), $b\equiv 1$ (mod 5), and $p\equiv 2$ (mod 5).
Suppose that $a<b$. Then we have that $a^{2}+a+1<p$.

###### Proof.

We again split into four cases and handle cases II and III via Lemma 20. In
Case I, we have that $x+a+1|p(a+b+1)+\frac{x-b}{b^{2}+b+1}$. We have that
$x+a+1\equiv 4$ (mod 5), and $p(a+b+1)+\frac{x-b}{b^{2}+b+1}\equiv 1$ (mod 5).
The rest of the case follows as usual.
For Case IV we have that $x-a|p(a+b+1)-\frac{x+b+1}{b^{2}+b+1}$. We have that
$x-a\equiv 4$ (mod 5) and that $p(a+b+1)-\frac{x+b+1}{b^{2}+b+1}\equiv 3$ (mod
5). The rest of the argument follows as usual. ∎

We are now in a position where we may prove:

######  Lemma 35.

There are no solutions to the equation

$x^{2}+x+1=(p^{2}+p+1)(q^{2}+q+1)(r^{2}+r+1)$

with $x$, $p$, $q$, $r$, $p^{2}+p+1$, $q^{2}+q+1$, $r^{2}+r+1$ all prime and
with $x\equiv pequiv1$ (mod 5) and $q\equiv 2$ (mod 5), and $r\equiv 3$ (mod
5). That is, there does not exist any triple threat $(x,p,q,r)$ where $x\equiv
p\equiv 1$ (mod 5) and $q\equiv 2$ (mod 5), $r\equiv 3$ (mod 5).

###### Proof.

Assume we have such an $(x,p,q,r)$. Either $p<q$ or $q<p$ (they cannot be
equal since they disagree mod 5). First, let us consider the case that $p<q$.
Then by Lemma 23 we have that $p<r$. We have then either $p<q<r$ or $p<r<q$.
Let us first consider the case of $p<q<r.$ We have then by Lemma 33 that $q<p$
which is a contradiction. Let us then consider the case $p<r<q$. We may then
use Lemma 34 to conclude that $p<r$ which is a contradiction. Thus, both of
the possibilities for $p<q$ lead to a contradiction. We thus must have $q<p$.
From $q<p$ and Lemma 24 we me must have $q<r$. Thus we have either $q<p<r$ or
$q<r<p$. If we have $q<p<r$, then by Lemma 34 we have that $p<q$. If $q<r<p$
we may use Lemma 34 to get that $r<q$ and so we have a contradiction. So in
each situation we have a contradiction and so the intended type of triple
threat does not exist. ∎

######  Lemma 36.

If $(x,a,b,c)$ is a triple threat with $x\equiv 1$ (mod 5) then none of $a,b$
or $c$ may be $1$ (mod 5).

###### Proof.

A check of possible triple threats mod 5 shows that (up to order of the
variables) every triple threat where $x\equiv 1$ (mod 5) and at least one of
$a$, $b$ or $c$ is $1$ (mod 5) then the triple threat must be one of a form
ruled out by Lemma 30, Lemma 35 or Lemma 25. ∎

######  Lemma 37.

Suppose that $a$ and $c$ are distinct odd primes, with $a^{2}+a+1$ prime.
Assume further that $a^{2}+a+1|c^{2}+c+1.$ Then $a^{2}+a+1<\frac{c}{2}$.

###### Proof.

Assume as given. Since $c^{2}+c+1\equiv 0$ (mod $a^{2}+a+1$), and $a^{2}+a+1$
is prime, we must have either $c\equiv a$ (mod $a^{2}+a+1)$ or $c\equiv a^{2}$
(mod $a^{2}+a+1$) (since $a^{2}+a+1|(c-a)(a+c+1)$). We $c\neq a$ by
assumption. We also have that $c\neq a^{2}$ since $c$ is prime. We then note
that $a+(a^{2}+a+1)$ and $a^{2}+(a^{2}+a+1)$ are both even and so $c$ cannot
be equal to either. Thus, we have that

$c\geq a+2(a^{2}+a+1)>2(a^{2}+a+1)$

from which the result follows. ∎

######  Lemma 38.

There are no odd primes $a,b,c$, with $a^{2}+a+1$ and $b^{2}+b+1$ prime, and
satisfying

$c^{2}+c+1=3(a^{2}+a+1)(b^{2}+b+1).$

###### Proof.

Assume we have odd primes $a,b,c$, with both $a^{2}+a+1$ and $b^{2}+b+1$
prime, and satisfying

$c^{2}+c+1=3(a^{2}+a+1)(b^{2}+b+1).$

We may then apply Lemma 37 twice to conclude that

$a^{2}+a+1<\frac{c}{2}$

and that

$b^{2}+b+1<\frac{c}{2}.$

We then have

$c^{2}+c+1=3(a^{2}+a+1)(b^{2}+b+1)<3\left(\frac{c}{2}\right)\left(\frac{c}{2}\right)=\frac{3c^{2}}{4}<c^{2}+c+1$

which is a contradiction. ∎

Similarly we have through nearly identical logic the following four results:

######  Lemma 39.

There are no solutions to $x^{2}+x+1=3(a^{2}+a+1)$ where $a,x$ and $a^{2}+a+1$
are all odd primes.

######  Lemma 40.

There are no solutions to $x^{2}+x+1=(a^{2}+a+1)(b^{2}+b+1)$ where $x,a,b$
$a^{2}+a+1$, and $b^{2}+b+1$ are all odd primes.

######  Lemma 41.

There are no solutions to $x^{2}+x+1=(a^{2}+a+1)(\frac{b^{2}+b+1}{3})$ where
$x,a,b$ $a^{2}+a+1$ and $\frac{b^{2}+b+1}{3}$ are all odd primes.

######  Lemma 42.

There are no solutions to
$x^{2}+x+1=(\frac{a^{2}+a+1}{3})(\frac{b^{2}+b+1}{3})$ where $x,a,b,$
$\frac{a^{2}+a+1}{3}$ and $\frac{b^{2}+b+1}{3}$ are all odd primes.

##  2 3 Divides N

We will in this section set $m=3^{\frac{f_{3}}{2}}ST$ where

$S=\prod_{p||m,p\neq 3}p$

and

$T=\prod_{p^{2}|m,p\neq 3}p.$

We will set $S=S_{1}S_{2}S_{3}S_{4}$ where a prime $p$ appears in $S_{i}$ for
$1\leq i\leq 3$ if $\sigma(p^{2})$ is a product of $i$ primes; $S_{4}$ will
contain all the primes of $S$ where $\sigma(p^{2})$ has at least 4 prime
factors. We will write $s=\omega(S)$ and write $t=\omega(T)$. We define
$s_{1}$, $s_{2}$ and $s_{3}$ similarly. We will write $S_{i,j}$ to be the
primes from $S_{i}$ which are $j$ (mod 3). In a similar way to use lower case
letters to denote the number of primes in each term as before and in general
will use a lower case letter to denote the number of distinct primes dividing
an upper case letter. Foe example,, we set $s_{i,j}=\omega(S_{i,j})$ and will
note that $s_{1,1}=0$. Thus, we do not need to concern ourselves with this
split for $S_{1}$ since all primes in $S_{1}$ are 2 (mod 3) there is no need
to split $S_{1}$ further. We will abuse notation slightly and will treat
capital letters as both products of distinct primes and as sets containing
those distinct primes. Thus will also think of lower case letters as denoting
the number of elements in the set formed by an upper case letter.

We have the special exponent is at least $1$:

$1\leq e$ (24)

We have the following straightforward equations from breaking down the
definitions of $s_{1},s_{2}$, $s_{3}$ and $s_{4}$:

$s=s_{1}+s_{2}+s_{3}+s_{4}.$ (25)

Similarly, we have:

$s_{2}=s_{2,1}+s_{2,2},$ (26) $s_{3}=s_{3,1}+s_{3,2},$ (27)

and

$s_{4}=s_{4,1}+s_{4,2}.$ (28)

We define $f_{4}$ as the number of prime divisors (counting multiplicity) in
$N$ which are not the special prime and are raised to at least the fourth
power. From simple counting we obtain:

$e+f_{3}+2s+f_{4}\leq\Omega.$ (29)

Due to Lemma 6 any element of $S_{3,1}$ must contribute at least one non-3
prime which is not from $S_{1}$. Motivated by this we will define $S_{3,1,T}$
to be the elements of $S_{3,1}$ which contribute two prime factors in $T$ or
$e$, and define $S_{3,1,S}$, as those which contribute two prime factors in
$S$. We will similarly define $S_{3,1,ST}$ as those which contribute one to
$S$ and one which goes to $T$ or $e$. We will similarly define $S_{1,T}$ as
the elements of $S_{1}$ which contribute to $T$ and define $S_{1,S}$ as the
elements of $S_{1}$ which contribute to $S$. Define $S_{1,e}$ as the set of
elements of $S_{1}$ which contribute the special prime. We will
correspondingly define $s_{3,1,T}$, $s_{3,1,S}$, $s_{3,1,ST}$ $s_{1,T}$,
$s_{1,S}$, and $s_{1,e}$. We of course have have $s_{1,e}\leq 1$ but we will
not need this here. We have then:

$s_{3,1}\leq s_{3,1,T}+s_{3,1,S}+s_{3,1,ST}.$ (30)

We then define $S_{3,1,{\bar{S}}T}$ as the set of elements of $S_{3,1,ST}$
which have their $S$ contributing term not arising from an $S_{1}$. Similarly
define $S_{3,1,S{\bar{T}}}$ as the set of elements of $S_{3,1,ST}$ which have
their $T$ term not arising from an $S_{1}$. We define as usual their lower
case variables for counting the number of elements in each set. From Lemma 6,
we have that

$S_{3,1,ST}=S_{3,1,{\bar{S}}T}\cup S_{3,1,S{\bar{T}}}.$

Thus, we have:

$s_{3,1,ST}\leq s_{3,1,{\bar{S}}T}+s_{3,1,S{\bar{T}}}.$ (31)

We also have

$s_{1}\leq s_{1,T}+s_{1,S}+s_{1,e}$ (32)

Here the $+1$ in the above inequality is due to the fact that we may have an
element which contributes the special prime.

######  Lemma 43.

We have

$s_{1}+s_{2,2}\leq t+s_{2,1}+s_{3,1}+s_{4,1}+1.$ (33)

###### Proof.

The proof of this lemma is essentially the same as that of Lemma 4 from [23].
∎

Next we have

$s_{2,1}+s_{3,1}+s_{4,1}\leq f_{3},$ (34)

since if $x\equiv 1$ (mod 3), then $x^{2}+x+1\equiv 0$ (mod 3).

We also have by counting all the $1$ (mod $3$) primes which are contributed by
primes in $S$

$s_{1}+2s_{2,2}+3s_{3,2}+s_{2,1}+2s_{3,1}+4s_{3,2}+3s_{3,1}\leq
f_{4}+e+2s_{2,1}+2s_{3,1}+2s_{4,1}.$

This simplifies to:

$s_{1}+2s_{2,2}+3s_{3,2}+4s_{4,2}+s_{4,1}\leq f_{4}+e+s_{2,1}.$ (35)

And we of course have

$4t\leq f_{4}.$ (36)

We may split $S_{2,2}$ into four sets, $S_{2,2,S}$,$S_{2,2,T}$,$S_{2,2,ST}$,
and $S_{2,2,e}$. We define $S_{2,2,e}$ as the elements of $S_{2,2}$ which
contribute to the special prime at least once. We define $S_{2,2,S}$ as the
elements of $S_{2,2}$ where both contribute to $S$, $S_{2,2,T}$ as the
elements of $S_{2,2}$ where both contribute to $T$, and $S_{2,2,ST}$ as the
elements of $S_{2,2}$ where one contributes to $S$ and one contributes to $T$.
We define their lower case variables as usual. We have:

$s_{2,2}\leq s_{2,2,S}+s_{2,2,T}+s_{2,2,ST}+s_{2,2,e}.$ (37)

We may split $S_{2,1}$ in a similar way into and $S_{2,1,S}$, $S_{2,1,T}$, and
$S_{2,1,e}$. We defined the lower case variables as usual. We have

$s_{2,1}\leq s_{2,1,S}+s_{2,1,T}+s_{2,1,e}.$ (38)

We have

$s_{1,e}+s_{2,1,e}+2*s_{2,2,e}\leq e.$ (39)

We have

$s_{1}\leq s_{1,S}+s_{1,T}+s_{1,e}.$ (40)

From Lemma 39, Lemma 40, Lemma 41 and Lemma 42 we get that every element of
$S_{2,2}$ must contribute at least one prime which does not arise from either
an $S_{1}$ or an $S_{2,1}$. Similarly, every element of $S_{2,2,ST}$ must have
either a contribution to $T$ or $e$ which does arise from an $S_{2,1}$ or
$S_{1}$ element or must have a contribution to $S$ which does not arise from
an $S_{2,1}$ or an $S_{1}$ element. We will set $S_{2,2,S*T}$ as those which
contribute an $S$ element of this form, and $S_{2,2,ST*}$ as those which
contribute a $T$ term. We define the lower-case counting variables as usual.
We have then:

$S_{2,2,ST}\leq S_{2,2,S*T}+S_{2,2,ST*}.$ (41)

We also have

$2s_{1,S}+2s_{2,1,S}+s_{2,2,S}+S_{2,2,S*T}+s_{3,1,S}+s_{3,1,\bar{S}T}\leq
2s_{2,1}+2s_{3,1}+2s_{4,1}.$ (42)

and

$4s_{1,T}+4s_{2,1,T}+s_{2,2,T}+s_{2,2,ST*}+s_{3,1,T}+s_{3,1,S\bar{T}}\leq
f_{4}+e$ (43)

We have from counting the primes from $S$ which are contributed by $S$

$s_{1,S}+s_{2,1,S}+2s_{2,2,S}+2s_{3,1,S}+s_{3,1,ST}\leq
2s_{2,1}+2s_{3,1}+2s_{4,1}.$ (44)

We have that

$s+t+2=\omega.$ (45)

Note that the $+2$ in Equation 45 arises from 3 and the special prime.

To prove the result we add up our inequalities as follows:
$\frac{9}{25}\times({\bf{\ref{specialexists3case}}})+\frac{16}{25}\times({\bf{\ref{sbreakdown}}})+\frac{16}{25}\times({\bf{\ref{s2breakdown}}})+\frac{16}{25}\times({\bf{\ref{s3breakdown}}})+\frac{16}{25}\times({\bf{\ref{s4breakdown}}})+1\times({\bf{\ref{bigomegalower2}}})+\frac{2}{25}\times({\bf{\ref{s31breakdown}}})+\frac{1}{25}\times({\bf{\ref{s
31st
breakdown}}})+\frac{8}{25}\times({\bf{\ref{s1breakdown}}})+\frac{2}{25}\times({\bf{\ref{s1s22upper}}})+1\times({\bf{\ref{3lower}}})+\frac{6}{25}\times({\bf{\ref{sbruteupper}}})+\frac{17}{25}\times({\bf{\ref{4tf4}}})+\frac{2}{25}\times({\bf{\ref{s22
breakdown}}})+\frac{8}{25}\times({\bf{\ref{s21
breakdown}}})+\frac{8}{25}\times({\bf{\ref{General contribution to special
prime}}})+\frac{1}{25}\times({\bf{\ref{star s and star t
breakdown}}})+\frac{7}{50}\times({\bf{\ref{star s
inequalty}}})+\frac{2}{25}\times({\bf{\ref{star t
inequality}}})+\frac{1}{25}\times({\bf{\ref{s to
s}}})+\frac{66}{25}\times({\bf{\ref{littleomegabreakdownwith3}}})$

##  3 3 does not divide N

For simplicity, we will prove the slightly weaker bound that

$\frac{302}{113}\omega-\frac{641}{113}\leq\Omega$ (46)

and then discuss the changes needed to improve the constant term. We set
$m=5^{\frac{f_{5}}{2}}11^{\frac{f_{11}}{2}}S^{2}T^{4}U^{\prime}$. Here we have

$S=\prod_{p,p||m,p\not\in\\{5,11\\}}p,T=\prod_{p,p^{2}||m,p\not\in\\{5,11\\}}p.$

We have $U=\frac{m}{5^{\frac{f_{5}}{2}}11^{\frac{f_{11}}{2}}S^{2}T^{4}}$ and
$U=rad(U^{\prime})$. That is,

$U=\prod_{p,p^{3}|m,p\not\in\\{5,11\\}}p.$

In other words, $S$ contains the prime divisors other than 5 and 11 which are
raised to exactly the second power in the factorization of $N$. Similarly, $T$
contains the prime divisors other 5, and 11 which are raised to exactly the
fourth power in the factorization of $N$. Finally, $U$ contains the prime
divisors other 5, and 11 and the special prime which are raised to at least
the sixth in the factorization of $N$. We set $s=\omega(S)$, $t=\omega(T)$ and
$u=\omega(U)$.

We have then

$\omega\leq s+t+u-3.$ (47)

We similarly define $f_{6}$ as the set of primes (counting multiplicity) who
appear to at least the 6th power. We have then:

$6u\leq f_{6}$ (48)

and

$e+f_{5}+2s+4t+f_{6}+f_{11}\leq\Omega.$ (49)

We do not have an equality in Equation 48 because there may be an element of
$u$ raised to a power higher than 6. We will define $S_{T}$ to be the elements
of $S$ which arise from $T$, that is

$S_{T}=\prod_{p|S,p|\sigma(T^{4})}p$

. Similarly, we will define $T_{S}$ to be the elements of $T$ which arise from
$S$.

$T_{S}=\prod_{p|T,p|\sigma(S^{2})}p.$

We note that any prime in $S_{T}$ must be $1$ (mod 5) or must be 5 itself. We
will define $S_{M}$ as the primes in $S$ which are $1$ (mod 5), and set
$s_{M5}=\omega(S_{M5})$. We will set $S=S_{1}S_{2}S_{3}S_{4}S_{5}$ similarly
to how we did in the case of $3|N$, with $p|S_{5}$ if $\sigma(p^{2})$ has 5 or
more not necessarily distinct prime factors, and similarly define
$s_{1},s_{2},s_{3},s_{4}$ and $s_{5}$. We then define $S_{M1},S_{M2}\cdots
S_{M5}$ as the intersections of the corresponding $S_{i}$ and $S_{M}$, and
then define $s_{M1},s_{M2}\cdots s_{M5}$ accordingly. We have

$s\leq s_{1}+s_{2}+s_{3}+s_{4}+s_{5}$ (50)

We have

$s_{M}\leq s_{M1}+s_{M2}+s_{M3}+s_{M4}+s_{M5}.$ (51)

We have that each of the $s_{i}$ is at least $s_{Mi}$ and thus we have:

$s_{M2}\leq s_{2}$ (52) $s_{M3}\leq s_{3}$ (53) $s_{M1}\leq s_{4}$ (54)
$s_{M5}\leq s_{5}$ (55)

Define $T_{S}$ to be the set of elements of $T$ which are contributed by $S$,
and define $t_{S}$ as the lower case variable as usual. Note that every
element of $T_{S}$ is 1 (mod 3). We note that

$t_{S}\leq t.$ (56)

We note that any prime factor contributed by $S$ cannot itself be in $S$. To
see why, note that any prime $p$ contributed by $S$ is $1$ (mod 3), and in
that case $3|\sigma(p^{2})$, but we are assuming in this section that
$3\not|N$. We have from Lemma 36 that every element of $S_{3M}$ must either
have all contributions be terms which do not arise from an $S_{1M}$ or must
contribute a term which does not arise from an $S_{1}$ at all. We will set
$S_{3MA}$ as the set of elements of $S_{3N}$ which contribute all terms not
arising from $S_{1M}$ and $S_{3MB}$ as the set of elements of $S_{3M}$ which
do not contribute at least one term which does not arise from $S_{1}$. We
define the lower case counting variables as usual. We have then

$s_{3M}\leq s_{3MA}+s_{3MB}.$ (57)

We have then

$4s_{1,m}+3s_{3MA}+s_{3MB}+s_{2}-3\leq 4t_{S}+u_{S}+e_{S}.$ (58)

Similarly we have

$4s_{1}+s_{3MB}+s_{2}-3\leq 4t_{S}+u_{S}+s_{e}.$ (59)

We also have by the same logic as Lemma 4 from [23].

$s_{1}+s_{2}\leq t_{S}+u_{S}+1.$ (60)

We also have from counting all the various $s_{i}$ contributions:

$s_{1}+2s_{2}+3s_{3}+4s_{4}+5s_{5}\leq 4t_{S}+u_{S}+e_{S}.$ (61)

We now turn to the equations which allow us to bound the number of primes in
$S_{M}$. To do this we need lower bounds on the contribution to $S$ from
elements in $T$. We define $T_{S1}$, $T_{S2}$, $T_{S3}$, $T_{S4}$ and $T_{S5}$
as follows: For $1\leq i\leq 4$ we define $T_{Si}$ to be the elements of
$T_{S}$ which contribute exactly $i$ primes, and we define $T_{S5}$ to be
those elements of $T_{S}$ which contribute at least 5 primes. We define
$T_{M}$ to be the set of elements of $T$ which are 1 mod 5.

$t_{S}\leq t_{S1}+t_{S2}+t_{S3}+t_{S4}+t_{S5}.$ (62)

We have

$t_{S1}+2t_{S2}+3t_{S3}+4t_{S4}+5t_{S5}\leq
2s_{M}+4t_{M}+u_{T}+e_{T}+f_{5}+f_{11}$ (63)

where $u_{t}$ and $e_{t}$ are defined analogously to $u_{s}$ and $e_{s}$.

We note that every element in $T_{S}$ is 1 (mod 3), and that if $x\equiv 1$
(mod 3), we have that $x^{4}+x^{3}+x^{2}+x+1\equiv 2$ (mod 3). But every
element of $S$ must be $1$ (mod 3). So if $p\in T_{S}$ we $\sigma(p^{4})\equiv
2$ (mod 3). Thus, any element of $T_{i,S}$ when $i$ is odd must contribute at
least one prime which is $1$ (mod 3 (and hence contribute a prime not in $S$),
since a product of an even number of numbers of the form $2$ (mod 3) will be
$1$ (mod 3). Thus we have:

$t_{S2}+t_{S4}\leq 4t_{M}+u_{t}+e_{t}.$ (64)

The next set of inequalities seeks to deal with the problem that we may have
very large $T_{S1}$. From Lemma 10 it follows that no element of $T_{S1}$ can
give rise to an element of $S_{1}$. For $i$ with $2\leq i\leq 5$ Define then
$S_{i*}$ as the elements of $S_{i}$ arising from $T_{1S}$. We define the lower
case counting variables as usual. We have then

$t_{S1}\leq s_{2*}+s_{3*}+s_{4*}+s_{5*}+t_{M}+u_{t}+e_{t}.$ (65)

We have

$s_{4*}\leq s_{4},$ (66)

and

$s_{5*}\leq s_{5}.$ (67)

We have from Lemma 8 and Lemma 9 that

$s_{1}+s_{3*}-1\leq t_{S}+u.$ (68)

We have from counting our contribution to the special prime that

$e_{S}+e_{T}\leq e.$ (69)

We also have

$u_{S}+u_{T}\leq f_{6}.$ (70)

We note that since every element in $T_{M}$ is 1 mod 5, we have

$t_{m}\leq f_{5}$ (71)

We also have by Lemma 10 that any contribution from $S_{2*}$ which contributes
both terms to $T$ must for a contribution to $f_{11}$. We set $S_{2}*T$ to be
those elements of $S_{2}*$ which contribute both terms to $T$, and we set
$S_{2*UE}$ to be those which contribute at least one term to either $u$ or
$e$. We have then:

$s_{2*}\leq s_{2*T}+s_{2*UE},$ (72) $s_{2*T}\leq f_{11},$ (73) $s_{2*UE}\leq
u-1.$ (74)

We have as before that the special prime must be raised to at least the first
power and thus:

$1\leq e.$ (75)

Finally we have that

$\omega\leq s+t+u+3.$ (76)

The 3 comes from the special prime, the possibility of division by 5 and the
possibility of division by 11.

To prove Inequality 46 we then add our inequalities as follows:
$\frac{302}{113}\times({\bf{\ref{little omega break down in (3,N)=1
case}}})+\frac{53}{113}\times({\bf{\ref{u bound
1}}})+1\times({\bf{\ref{bigomegabreakdown no
3s}}})+\frac{76}{113}\times({\bf{\ref{s breakdown for (3,N)=1
case}}})+\frac{8}{113}\times({\bf{\ref{SM
breakdown}}})+\frac{8}{113}\times({\bf{\ref{s2 is at least
sm2}}})+\frac{2}{113}\times({\bf{\ref{s3 is at least
sm3}}})+\frac{8}{113}\times({\bf{\ref{s4 is at least
sm4}}})+\frac{8}{113}\times({\bf{\ref{s5 is at least
sm5}}})+\frac{150}{113}\times({\bf{\ref{t_s lower bound for
t}}})+\frac{6}{113}\times({\bf{\ref{s3m
breakdown}}})+\frac{2}{113}\times({\bf{\ref{s3ma
bound1}}})+\frac{4}{113}\times({\bf{\ref{s3mb
bound1}}})+\frac{26}{113}\times({\bf{\ref{s1 + s2 for 3 not dividie N
case}}})+\frac{26}{113}\times({\bf{\ref{brute s_i contribution for 3 not
divide N}}})+\frac{12}{113}\times({\bf{\ref{T
breakdown}}})+\frac{4}{113}\times({\bf{\ref{ts
brute}}})+\frac{4}{113}\times({\bf{\ref{t_even
contribution}}})\frac{8}{113}\times({\bf{\ref{ts1 gives rise to miraculous
factorizations}}})+\frac{8}{113}\times({\bf{\ref{s4* <
s4}}})+\frac{8}{113}\times({\bf{\ref{s5* <
s5}}})+\frac{8}{113}\times({\bf{\ref{s_1 +
s_3*}}})+\frac{32}{113}+\times({\bf{\ref{es + et <=
e}}})\frac{58}{113}\times({\bf{\ref{f_6 lower
bound}}})+\frac{40}{113}\times({\bf{\ref{t_m to
f_5}}})++\frac{8}{113}\times({\bf{\ref{s_2*
breakdown}}})+\frac{8}{113}\times({\bf{\ref{s_2* to T forces
11}}})+\frac{8}{113}\times({\bf{\ref{s_2*UE
contribution}}})+\frac{81}{113}\times({\bf{\ref{special prime exists in 3 not
divide N case}}})$ It follows from the results in [4] that if one has any of
$5^{2}||N$, $5^{4}||N$, $11^{2}||N$, or $11^{4}||N$ then one must have at
least one prime which is not the special prime raised to at least the 6th
power, and hence have $u\geq 1$. We thus may adjust the above equations
slightly: If $(N,55)=1$, we have instead of 47, we have $\omega=s+t+u-1$ and
$f_{5}=f_{11}=0$. In this case we get a bound of

$\frac{302}{113}\omega-\frac{286}{113}\leq\Omega.$ (77)

Alternatively, one must have $5^{6}|N$ or $11^{6}|N$ if either $5|N$ or
$11|N$. We can without too much work check that all of these force a bound at
least as tight as 77. Thus we always have that bound.

##  4 Improved Norton type results

Norton [16] proved two types of results. First, he proved lower bounds for
$\omega(N)$ in terms of the smallest prime factor of $N$. Second, he proved
lower bounds for $N$ in terms of its smallest prime factor. In this section,
we will slightly improve Norton’s first type of result and show how we can
combine that with the Ochem-Rao type results to substantially improve the
second type of result. Set $P_{n}$ to be the $n$th prime number. For $n>1$
Norton defined $a(n)$ as the integer such that

$\prod_{r=n}^{n+a(n)-2}\frac{P_{r}}{P_{r}-1}<2<\prod_{r=n}^{n+a(n)-1}\frac{P_{r}}{P_{r}-1}.$
(78)

It is easy to see that if $N$ is an odd perfect number with smallest prime
divisor $P_{n}$, then $N$ must have at least $a(n)$ distinct prime divisors.
In fact, although Norton does not state this explicitly, this statement also
applies to any odd abundant number $N$. Thus, study of $a(n)$ is a natural
object even if one is not strongly interested in odd perfect numbers. Norton
proved

######  Theorem 44.

(Norton)[16] Let $N$ be an odd perfect number with smallest prime divisor
$P_{n}$, largest prime divisor $P_{s}$, and let $b$ be a constant less than
$\frac{4}{7}$. Then we have:

  1. 1.

$a(n)=Li(P_{n}^{2})+O\left(n^{2}e^{-\log^{b}n}\right).$ (79)

  2. 2.

$a(n)=\frac{1}{2}n^{2}\log n+\frac{1}{2}n^{2}\log\log
n-\frac{3}{4}n^{2}+\frac{n^{2}\log\log n}{2\log n}+O\left(\frac{n^{2}}{\log
n}\right)$ (80)

  3. 3.

$P_{s}\geq P_{n+a(n)-1}=P_{n}^{2}+O\left(n^{2}e^{-\log^{b}n}\right).$ (81)

  4. 4.

$P_{s}\geq P_{n+a(n)-1}=n^{2}\log^{2}n+2n^{2}\log n\log\log n-2n^{2}\log
n+n^{2}(\log\log)^{2}+O(n^{2})$ (82)

  5. 5.

$\log N>2P_{n}^{2}+O\left(n^{2}e^{-\log^{b}n}\right).$ (83)

Note that only Equation 83 is using non-trivial material about odd perfect
numbers. The bounds for $P_{s}$ apply to any odd abundant number, and the
bounds for $a(n)$ do not depend on odd perfect numbers at all. These bounds of
Norton are not by themselves constructive; Norton proved slightly weaker
constructive bounds.333Prior to Norton a similar non-constructive but more
general bound was proven which gives a lower bound for the $\omega(n)$ in
terms of $\alpha$ and $p$ where $n$ satisfies $\frac{\sigma(n)}{n}\geq\alpha$
and $n$ has smallest prime factor $p$.[20]

######  Theorem 45.

(Norton)[16] Let $N$ be an odd perfect number with smallest prime divisor
$P_{n}$, and set $P_{s}=P_{a(n)-n-1}$. Then we have

  1. 1.

$a(n)>n^{2}-2n-\frac{n+1}{\log n}-\frac{5}{4}-\frac{1}{2n}-\frac{1}{4n\log
n}.$ (84)

  2. 2.

As long as $n\geq 9$,

$\log N>2P_{s}\left(1-\frac{1}{2\log
P_{s}}\right)-2P_{n}\left(1+\frac{1}{2\log P_{n}}\right)+6\log P_{n}+2\log
P_{n+1}-\log P_{s}.$ (85)

These bounds are explicit, with the cost of being weaker in form than the
bounds in Theorem 44. To prove Equation 84 Norton did have to use results
about odd perfect numbers. In fact, Norton’s method uses some early results
ruling out specific forms of odd perfect numbers, although the forms ruled out
only allow improvement in the lower order terms of his inequality. Subsequent
results in this section can be thought of as using similar ideas, but because
we have stronger results on what an odd perfect number can look like, we can
actually improve the constant in front of the lead terms. It is also worth
noting that for large values of $n$, Norton’s explicit bound for $a(n)$ give a
better result than the often cited bound of Grun [7], which shows that if $N$
is an odd perfect number with smallest prime divisor $p$, then

$(3/2)p-2\leq\omega(N).$ (86)

Norton’s result focus on $a(n)$, but for some purposes it is more natural to
look at the function $b(p)$, defined for odd primes $p$, where $b(p)=a(n)$
where $p=p_{n}$. We will examine the behavior of both functions in this
section. While Norton’s results in work for abundant or perfect numbers, we
will also be interested in how they can be strengthened for odd perfect
numbers. In this context, we will define $b_{o}(p)$ to be the minimum of the
number of distinct prime divisors of any odd perfect number with smallest
prime divisor $p$. We will set $b_{o}(p)=\infty$ when there are no odd perfect
numbers with smallest prime divisor $p$. We will define $a_{o}(n)$ in
analogous fashion. Trivially one has $b_{o}(p)\geq b(p)$ and $a_{o}(n)\geq
a(n)$. One would like to be able to prove that $b_{o}(p)>b(p)$ but this seems
very difficult.

In this section, we will do three things: First, we will prove strengthened
versions of Norton’s constructive results bounding $a(n)$ from below and a
similar one for $b(p)$ although they will still fall slightly short of the
non-constructive results. Second, we will construct a general framework to use
Ochem and Rao type results to get explicit inequalities similar to 85 which in
general are better than Norton (both his explicit form and his non-
constructive form in Equation 83). Third, we will use this framework and our
earlier results to construct a strong lower bound for the size of an odd
perfect number in terms of its smallest prime factor. We will write
$S(x)=\prod_{p\leq x}\frac{p}{p-1}.$ We will write $\vartheta(x)$ to be
Chebyshev’s first function, that is

$\vartheta(x)=\sum_{p\leq x}\log p.$

We need the following lemma:

######  Lemma 46.

For any prime $p>2$, we have $b(p)\geq p$.

###### Proof.

Assume that $b(p)=m$. Then since $\frac{x}{x-1}$ is a decreasing function for
positive $x$ and the $i$th prime after $p$ is at least $p+i$, we must have

$2<\left(\frac{p}{p-1}\right)\left(\frac{p+1}{p}\right)\left(\frac{p+2}{p+1}\right)\cdots\left(\frac{p+b(p)-1}{p+b(p)-2}\right)=\frac{p+b(p)-1}{p-1}.$

Thus

$2p-2<p+b(p)-1,$

and so $b(p)>p-1$, and so $b(p)\geq p$. ∎

Note that variants of the above lemma are very old. Often this result is
normally stated simply that for an odd perfect number needing to at least as
many distinct prime factors than their smallest prime divisor. The lemma when
stated applying just to perfect numbers seems to back to Servais[21] and has
been proved repeatedly such as in [15]. In that context, it is worth noting
that there is also a slightly stronger version in the literature which again
has been proven a few times. In this case, the oldest version appears to be
due to Grun [7]. As previously discussed, Grun proved that if $N$ is an odd
perfect number, with least prime divisor $p$, then $\frac{2}{3}\omega(N)+2\geq
p$. Again, the proof can, with no substantial effort, be generalized to a
statement about $b(p)$. in Grun’s proof, the key observation that odd primes
must differ by at least 2, and therefore can instead use the inequality

$2<\left(\frac{p}{p-1}\right)\left(\frac{p+2}{p+1}\right)\left(\frac{p+4}{p+3}\right)\cdots\left(\frac{p+2b(p)-2}{p+b(p)-3}\right),$

and then estimate this quantity. As with the lemma of Servais, Grun’s lemma
applies to any odd perfect or odd abundant number although it is normally
phrased simply for odd perfect numbers.

We will need a few explicit estimates of certain functions of primes. We have
[3] that

$\frac{x}{\log x}\left(1+\frac{0.992}{\log
x}\right)\leq\pi(x)\leq\frac{x}{\log x}\left(1+\frac{1.2762}{\log x}\right),$
(87)

with the lower bound valid if $x\geq 599$ and the upper bound valid for all
$x>1$. We also have for $n\geq 2$,

$P_{n}\geq n\left(\log n+\log\log n-1+\frac{32}{31(\log n)^{2}}\right).$ (88)

The above bound for $P_{n}$ follows from the following bound in [3] which
differs only in the last term:

$P_{n}\geq n\left(\log n+\log\log n-1+\frac{\log\log x-9/4}{\log x}.\right).$
(89)

To obtain 89, we note that for if $n\geq 35312$, we have that

$\frac{\log\log n-9/4}{\log n}\geq\frac{32}{31(\log n)^{2}},$

and we may then verify the inequality for all $n$ with $2\leq n\leq 35312$. We
will write

$\underline{P}_{n}=n\left(\log n+\log\log n-1+\frac{32}{31(\log
n)^{2}}\right).$

######  Lemma 47.

If $A$ and $B$ are real numbers, and $A\geq B\geq 6$ satisfying

$A+\frac{1}{2A}\geq 2B-\frac{1}{B},$

then

$A\geq 2B-\frac{5}{4B}-\frac{1}{37B^{2}}.$

###### Proof.

Note that $A+\frac{1}{2A}$ is a positive increasing function in $A$, to prove
this one simply needs verify that if $B\geq 6$ then

$2B-\frac{5}{4B}-\frac{1}{37B^{2}}+\frac{1}{2(2B-\frac{5}{4B}-\frac{1}{37B^{2}})}\leq
2B-\frac{1}{B}.$

∎

For $x>1$ we have [19]

$e^{\gamma}(\log x)\left(1-\frac{1}{2\log^{2}x}\right)<S(x)<e^{\gamma}(\log
x)\left(1+\frac{1}{\log^{2}x}\right)$ (90)

where $\gamma$ is Euler’s constant. We will write

$Q(x,y)=\prod_{y<p\leq x}\frac{p}{p-1}=\frac{S(x)}{S(y)}$

and will assume that $x>y$. We have from Equation 90 that

$\frac{\log x}{\log
y}\left(\frac{1-\frac{1}{2\log^{2}x}}{1+\frac{1}{\log^{2}y}}\right)<Q(x,y)<\frac{\log
x}{\log y}\left(\frac{1+\frac{1}{\log^{2}x}}{1-\frac{1}{2\log^{2}y}}\right).$
(91)

######  Theorem 48.

For all $n>1$ we have

$a(n)\geq\frac{n^{2}}{2}\left(\log n-\frac{3}{2}\log\log
n+\frac{1}{20}+\frac{\log\log n}{\log n}\right)-n+1.$

###### Proof.

We may verify from direct computation that the above inequality is valid for
$n\leq 117$, and so we may assume that $n\geq 118$ (equivalently that $p\geq
647$). Our plan to estimate $a(n)$ is to set $y=P_{n}$ in Equation 91, find a
lower bound on $x$ such that $Q(x,y)>2$, and then estimate $\pi(x)$ since we
will have $a(n)=\pi(x)-n+1$. We have from Equation 91 that

$2<\frac{\log x}{\log
y}\left(\frac{1+\frac{1}{\log^{2}x}}{1-\frac{1}{2\log^{2}y}}\right)$

which implies that

$\log x+\frac{1}{2\log x}\geq 2\log y-\frac{1}{\log y}.$

We have then from Lemma 47 that

$\log x\geq 2\log y-\frac{5}{4\log y}-\frac{1}{37(\log y)^{2}}.$ (92)

(In the above use of the Lemma we are setting $A=\log x$ and $B=\log y$.) Note
that getting a lower bound $a(n)$ is exactly the same as lower bounding the
minimum $x$, such that $Q(x,p_{n})>2$, and then getting a lower bound on
$\pi(x)-\pi(y)=\pi(x)-n+1$. It is not hard to see that Equation 92 yields that
as as long as $y\geq 541$ that

$x\geq y^{2}\left(1-\frac{5}{4\log y}+\frac{1}{2(\log y)^{2}}\right).$ (93)

We now need to estimate $a(n)$ by estimating $\pi(x)-n+1$. We have then from
Equation 93 and Equation 89, along with the fact that the function
$j(s)=1-\frac{5}{4s}+\frac{1}{2s^{2}}$ is increasing for $t\geq 1/10$ that

$a(n)\geq\pi\left((\underline{P}_{n})^{2}\left(1-\frac{5}{4\log\underline{P}_{n}}+\frac{1}{2(\log\underline{P}_{n})^{2}}\right)\right)-n-1$
(94)

We have the trivial estimate that $\underline{P}_{n}\geq n$, and so we have
again using that $f(s)$ is increasing in $s$, and substituting in the
definition of $\underline{P}_{n}$, and set $t=\log n$ that

$a(n)\geq\pi\left(n^{2}\left(t+\log
t-1+\frac{32}{31t^{2}}\right)^{2}\left(1-\frac{5}{4t}+\frac{1}{2t^{2}}\right)\right)-n-1.$
(95)

When $t>4.77$, one has that

$\left(t+\log
t-1+\frac{32}{31t^{2}}\right)^{2}\left(1-\frac{5}{4t}+\frac{1}{4t^{2}}\right)\geq
t^{2}-\frac{\log t}{2}.$

So for $n$ in the range under discussion we have

$a(n)\geq\pi\left(n^{2}(t^{2}-\frac{\log
t}{2})\right)-n+1=\pi\left(n^{2}((\log n)^{2}-\frac{\log\log
n}{2})\right)-n+1.$ (96)

We wish to apply Equation 87 to 96. To do so, we need a lower bound on

$\frac{1}{\log\left(n^{2}((\log n)^{2}-\frac{\log\log n}{2})\right)}.$

It is not hard to check that as long as $n>e^{e}$ one has that

$\frac{1}{\log\left(n^{2}((\log n)^{2}-\frac{\log\log
n}{2})\right)}\geq\frac{1}{2\log n}\left(1-\frac{\log\log n}{\log n}\right).$
(97)

We can now apply Equation 97 to Equation 96 and 87 to get that

$a(n)\geq\frac{n^{2}\left(t^{2}-\frac{t\log t}{2}\right)\left(1-\frac{\log
t}{t}\right)}{2t}\left(1+.992\left(\frac{1}{2t}-\frac{\log
t}{2t^{2}}\right)\right)-n-1.$ (98)

We have again in the above for convenience written $\log n$ as $t$. A little
work then shows that for $n\geq 43$, we have that the right-hand side of
Equation 98 is at least

$\frac{n^{2}}{2}\left(\log n-\frac{3}{2}\log\log n+\frac{1}{20}+\frac{\log\log
n}{\log n}\right)-n+1$

which proves the theorem. ∎

Note that the bound given by Theorem 48 is tighter than the bound from
Equation 84. We similarly have an interest in estimating $b(p)$. We have then:

######  Theorem 49.

For all primes $p>2$, we have

$b(p)\geq\frac{p^{2}}{2\log p}\left(1-\frac{0.754}{\log p}-\frac{0.745}{(\log
p)^{2}}-\frac{0.247}{(\log p)^{3}}+\frac{0.631813}{(\log
p)^{4}}\right)-\pi(x)+1.$ (99)

and

$b(p)\geq\frac{p^{2}}{2\log p}\left(1-\frac{0.754}{\log p}-\frac{745}{(\log
p)^{2}}-\frac{0.247}{(\log p)^{3}}+\frac{0.631813}{(\log
p)^{4}}\right)-\frac{p}{\log p}(1+\frac{1.2726}{\log p})+1.$ (100)

###### Proof.

We will prove only the second of the two inequalities (the proof for the first
statement is nearly identical). The first few steps in this proof are
essentially identical to those in the proof of Theorem 48. We again assume
that $p\geq 647$, and proceed until we reach Equation 93. And as before we
estimate

$\pi(x)-n+1.$

We need to lower bound the left-hand side of

$b(p)\geq\pi\left(p^{2}\left(1-\frac{5}{4\log p}+\frac{1}{2(\log
p)^{2}}\right)\right)-\pi(p)+1.$ (101)

We need to apply 87. We note that although the lower bound on 87 requires that
the argument of $x$ be at least 599, we have that in this case since $p\geq
647$. We also need a lower bound estimate for

$\frac{1}{\log(p^{2}(1-\frac{5}{4\log p}+\frac{1}{2(\log p)^{2}})}.$

It is not hard to verify that when $p$ in our range we have

$\frac{1}{\log(p^{2}(1-\frac{5}{4\log p}+\frac{1}{2(\log
p)^{2}})}\geq\frac{1}{2\log p}\left(1-\frac{5}{8(\log
p)^{2}}-\frac{21}{32(\log p)^{3}}\right).$ (102)

We will set $t=\log p$, and then use Equation 102 to apply Equation 87 to 101
to get that:

$b(p)\geq\frac{p^{2}}{2t}\left(1-\frac{5}{4t}+\frac{1}{2t^{2}}\right)\left(1-\frac{5}{8t^{2}}-\frac{21}{32t^{3}}\right)\left(1+\frac{0.992}{2t}\left(1-\frac{5}{8t^{2}}-\frac{21}{32t^{3}}\right)\right)-\frac{p}{t}(1+\frac{1.2726}{t})+1.$
(103)

We need then to estimate

$I_{2}(t)=\left(1-\frac{5}{4t}+\frac{1}{2t^{2}}\right)\left(1-\frac{5}{8t^{2}}-\frac{21}{32t^{3}}\right)\left(1+0.992\left(1-\frac{5}{8t^{2}}-\frac{21}{32t^{3}}\right)\right).$

We have

$I_{2}=1-\frac{0.754}{t}-\frac{0.745}{t^{2}}-\frac{0.247}{t^{3}}+\frac{0.631813}{t^{4}}+E(t)$

Where

$E(t)=\frac{0.369375}{t^{5}}-\frac{0.160813}{t^{6}}-\frac{0.198109}{t^{7}}-\frac{0.0635742}{t^{8}}+\frac{0.106805}{t^{9}}.$

We note that $E(t)$ is positive when $t>1$ which is satified in our range.
Thus we conclude that

$I_{2}(t)\geq
1-\frac{0.754}{t}-\frac{0.745}{t^{2}}-\frac{0.247}{t^{3}}+\frac{0.631813}{t^{4}}$

which proves the theorem. ∎

Note that although Theorem 49 and Theorem 48 both give the asymptotically
correct values, in practice Theorem 49 is stronger. This is due to Theorem 48
requiring that we use a lowest bound estimate for $P_{n}$ in terms of $n$.

While we cannot directly show that $b_{o}(n)>b(p)$, we can get partial results
of this form. In particular, $b(p)=3$, but $b_{o}(3)\geq 10$ [12]. Similarly,
$b(5)=7$, and $b_{o}(5)\geq 12$ [10]). We will also prove a similar result for
other small values of $b(p)$ using the fact that the largest prime divisor of
an odd perfect number must be at least $10^{8}$ [6].

######  Proposition 50.

We have $b_{o}(p)\geq b(p)+1$, for $p\leq 397$.

###### Proof.

We will show the calculation for $p=7$. The calculation is nearly identical
for the other primes in question. We note that $b(7)=15.$ Now, assume $N$ is
an odd perfect number with smallest prime divisor 7, and with exactly 15
distinct prime divisors. Then we have

$2=\frac{\sigma(N)}{n}<H(n)\leq\frac{7}{6}\frac{11}{10}\frac{13}{12}\frac{17}{16}\cdots\frac{53}{52}\frac{59}{58}\frac{10^{8}+1}{10^{8}}<1.994.$

This is a contradiction. ∎

This proposition stops at 397 is because the relevant product is actually
greater than 2 for the next prime, 401. The result could be extended if the
result from [6] could be extended further; however, extending that result (say
to that an odd perfect number must have a prime divisor which is at least
$10^{9}$) would likely take either very heavy new computations or would take
some fundamental new insight. That said, it is plausible that a similar result
could be proved just for odd perfect numbers not divisible by any prime less
than some bound, and this would allow one to extend the above proposition in
this case. We also have as a consequence of Theorem 49:

######  Corollary 51.

If $p\geq 11$, then $b(p)\geq 2p+2$.

Using Proposition 50, Corollary 51 and the earlier remarks for $p=3$ and
$p=5$, we can combine this with Theorem 49 bound to obtain with a little
work::

######  Corollary 52.

Let $N$ be an odd perfect number with smallest prime factor $q$. Then we have
$\omega(N)\geq 2q+2$.

###### Proof.

The result is essentially just Corollary 51 except for $p=3,5,7$. $p=3$ is
handled since an odd perfect number must have at least 10 distinct prime
factors and $10\geq 2(3)+2$. Since an odd perfect number not divisible by 3
must have at least 12 prime factors, 5 is likewise handled. Since $b(7)=15$,
we have that $b_{o}(p)\geq 16$ by Proposition 50. And so the result is proven.
∎

Note that Corollary 52 is tighter than Grun’s result for all odd primes $p$.
One can see from examples like $945$ that this bound really does require that
$N$ is an odd perfect number, unlike Grun’s bound which applies also to odd
abundant numbers. It is easy to see from the definition of $a(n)$ that
$a(n+1)\geq a(n)$ for all $n\geq 2$. However, Norton’s bounds do not appear by
themselves to be tight enough to conclude that $a(n+1)>a(n)$ for all $n\geq
2$. But we can use Corollary 51 to prove this result.

######  Proposition 53.

For all $n\geq 2$ We have $a(n+1)\geq a(n)+1$. Equivalently, if $P_{n}$ is an
odd prime and $P_{n+1}$ is the next prime after $P_{n}$ then $b(P_{n+1})\geq
b(P_{n})+1$.

###### Proof.

We can verify that the statement is true for any prime $p\leq 17$, so we may
without loss of generality assume that $P_{n+1}>P_{n}\geq 19$. Assume that
$b(P_{n})=b(P_{n+1})=m$, or equivalently that $P_{n}=P_{n+1}=m$. This means we
have that

$\prod_{r=n}^{n+m-2}\frac{P_{r}}{P_{r}-1}<2<\prod_{r=n}^{n+m-1}\frac{P_{r}}{P_{r}-1}.$
(104)

and

$\prod_{r=n+1}^{n+m-1}\frac{P_{r}}{P_{r}-1}<2<\prod_{r=n+1}^{n+m}\frac{P_{r}}{P_{r}-1}.$
(105)

We have then

$\prod_{r=n+1}^{n+m}\frac{P_{r}}{P_{r}-1}=\left(\prod_{r=n}^{n+m-2}\frac{P_{r}}{P_{r}-1}\right)\left(\frac{P_{n}-1}{P_{n}}\right)\left(\frac{P_{n+m-1}}{P_{n+m-1}-1}\right)\left(\frac{P_{n+m}}{P_{n+m-1}-1}\right).$
(106)

However, we have from Equation that 104 the first term on the right hand-side
of Equation 106 is less than 2. We claim that the remaining terms are less 1,
which would mean that the right-hand side of Equation 105 would be both
greater than 2 and less than 2 which is a contradiction. It just remains to
show that

$\left(\frac{P_{n}-1}{P_{n}}\right)\left(\frac{P_{n+m-1}}{P_{n+m-1}-1}\right)\left(\frac{P_{n+m}}{P_{n+m-1}-1}\right)<1.$

We note that $b(P_{n})\geq 2P_{n}+2$, and thus $P_{n+m_{1}}\geq 2p+1$. We have
then $P_{n+m}\geq 2P_{n}+3$. Thus we have that

$\left(\frac{P_{n}-1}{P_{n}}\right)\left(\frac{P_{n+m-1}}{P_{n+m-1}-1}\right)\left(\frac{P_{n+m}}{P_{n+m-1}-1}\right)\leq\left(\frac{x-1}{x}\right)\left(\frac{2x+1}{2x}\right)\left(\frac{2x+3}{2x+2}\right),$

where $P_{n}=x.$ However, we have that

$\left(\frac{x-1}{x}\right)\left(\frac{2x+1}{2x}\right)\left(\frac{2x+3}{2x+2}\right)=\frac{4x^{3}+4x^{2}-5x-3}{4x^{3}+4x^{2}}<1.$

This completes the proof. ∎

In a similar vein, one can easily modify the above proof to conclude:

######  Proposition 54.

For any constant $c$ there are only finitely many $n$ where $a(n+1)-a(n)\leq
c$.

We have just shown that $a(n)$ is increasing in $n$. This is the same as
saying that the first difference, $a(n+1)-a(n)$, is always positive. One might
naturally wonder about the behavior of the second differences. Since $a(n)$
asymptotically behaves like $\frac{1}{2}n^{2}\log n$ which has positive and
indeed slightly increasing second differences, one might hope that $a(n)$ at
least has always positive second differences. Alas, this is not the case. Let
$f(n)$ be the second difference of $a(n)$, that is,
$f(n)=a(n+1)+a(n-1)-2a(n)$. Generally, $f(n)$ is positive. However,
$f(31)=-5$, and $f(100)=-144$. What is going on here? The key issue appears to
be that both of these values correspond to primes which occur right after a
large gap. We say that a prime $P_{n}$ occurs after a record setting gap if
$P_{n}-P_{n-1}$ is larger than it is for any other choice of smaller $n$. In
particular, the 30th prime is 113, and then there is a record-setting gap to
the 31st prime of 127. Similarly, the 99th prime is 523 and then there is a
record setting gap until the 100th prime of 541. This shouldn’t be surprising.
Because there are unusual gaps here, $a(30)$ and $a(99)$ need to be extra
large since the relevant products lack any smallish primes other than $P_{30}$
and $P_{99}$. (Remember the smaller a prime the more it contributes to our
product.) We can check this intuition by looking at when $f(n)=1$ and noting
the two smallest examples of this occur at $n=10$, corresponding to the record
setting gap between 23 and 29, and at $n=25$, corresponding to the record
setting gap between 83 and 89. Note that we can have $f(n)=1$ when the gap is
not a record setting gap, such as at $n=35$, which corresponds to the large
but not record setting gap between 139 and 149. This discussion leads to four
questions about the behavior of $f(n)$ and one about $a(n+1)-a(n)$.

  1. 1.

Are there infinitely many values of $n$ where $f(n)$ is negative?

  2. 2.

Is the set of $n$ where $f(n)<0$ a subset of those $n$ where $P_{n}$ is right
after a record setting gap?

  3. 3.

Are there infinitely many $n$ where $P_{n}$ occurs after record setting gaps
and $f(n)$ is positive?

  4. 4.

Does $f(n)$ take on any integer value? In particular, is $f(n)$ ever zero?

  5. 5.

Does $a(n+1)-a(n)$ take on every positive integer value?

##  5 Hybrid bounds

We wish to combine the Norton type results together with the Ochem-Rao type
results to get a strong lower bound on the size of an odd perfect number in
terms of its smallest prime factor. We will write $b_{2}(p)=b(p)+\pi(p)-1$.

Let $S$ be a set of odd primes. We say $N$ is an _S-avoiding OPN_ if $N$ is an
odd perfect number not divisible by any prime in $S$. Notice in particular
that if the smallest prime factor of $N$ is $p$, then $N$ is an $S$ avoiding
OPN with $S$ the set of odd primes strictly less than $p$. Given $S$ a set of
primes (possibly empty), and $\alpha$, and $\beta$ to be real numbers, we will
write OR($\alpha$, $\beta$, $S$) for the statement “For any $S$-OPN, we have
$\Omega(N)\geq\alpha\omega+\beta$.” In this framework, Ochem and Rao’s
original result of Equation1 is the statement OR($\frac{18}{7}$,
$\frac{-31}{7}$, $\emptyset$). Similarly, Equations 3 and 4 can be stated as
OR($\frac{8}{3}$,$\frac{-7}{3}$, $\\{3\\}$) and OR($\frac{21}{8}$,
$\frac{-39}{8}$, $\emptyset$). Theorem 2 can be stated
OR($\frac{302}{113}$,$\frac{-286}{113}$,$\\{3\\}$) and
OR($\frac{66}{25}$,$-5$,$\emptyset$).

######  Theorem 55.

Let $S$ be a finite set of odd primes. Let $\alpha$ and $\beta$ be real
numbers with $\alpha>2$. Let $M$ be the maximum of $S$. Assume that $p>M$. Let
$N$ be an odd perfect number with smallest prime factor $p$, and also
satisfying $\alpha\omega(N)+\beta\geq 0)$. Set $Q=P_{n+b(p)-1}$. Then we have:

$\log N\geq(\log
p)\left((\alpha-2)(b(p))-\beta+1\right)+2(\vartheta(Q)-\vartheta(p))-\log Q.$

###### Proof.

Assume as given and note that every prime factor of an odd perfect number
except possibly the special prime must be raised to at least the second power.
This contributes the $2(\vartheta(P)-\vartheta(p))-\log Q$ term (where in the
worst case scenario $Q$ is the special prime). However, we have an additional
contribution of the remaining primes which are forced by our lower bound for
$\Omega(N)$. Each of those primes is at least $p$, and there are at least
$((\alpha-2)\omega(N)+\beta+1$ such primes (with the +1 coming from our
special prime only being raised to the first power rather than the second).
This give us the other term above. ∎

We will need two following results from [19].

$x\left(1-\frac{1}{2\log x}\right)<\vartheta(x)<x\left(1+\frac{1}{2\log
x}\right).$ (107)

Here $\vartheta(x)$ is Chebyshev’s second function, that is
$\vartheta(x)=\sum_{p\leq x}\log p$, and the upper bound is valid for $x>563$
and the lower bound is valid for $x>1$. We have as an immediate corollary of
Equation 99

######  Corollary 56.

Let $N$ be an odd perfect number with smallest prime factor $p$. Then we have

$b_{2}(p)\geq\frac{p^{2}}{2\log p}\left(1-\frac{0.754}{\log
p}-\frac{0.745}{(\log p)^{2}}-\frac{0.247}{(\log p)^{3}}+\frac{0.631813}{(\log
p)^{4}}\right)$ (108)

We can use this sort of result to get results stronger than Norton’s lower
bounds for $\log N$ in Equation 83 Equation 85. We have using our previous
bounds and a little algebra the following:

######  Lemma 57.

Let $p$ be an odd prime greater than 3 . Set $t=\log p$ Then

$P_{b_{2}(p)}\geq p^{2}I_{3}(t)$ (109)

where

$I_{3}(t)=1-\frac{0.754}{t}-\frac{2.5\log
t}{t^{2}}-\frac{1.808}{t^{2}}-\frac{0.55\log t}{t^{3}}+\frac{0.41(\log
t)^{2}}{t^{4}}+\frac{0.2\log t}{t^{4}}+\frac{3.6}{t^{4}}.$ (110)

We can use Lemma 55 and Lemma 110 with our bound for $b(p)$ from Theorem 49 as
well as the statement OR($\frac{8}{3}$,$\frac{7}{3}$, $\\{3\\}$) and Ochem and
Rao’s bound that $N>10^{1500}$ to obtain:

######  Theorem 58.

Let $N$ be an odd perfect number number with smallest prime divisor $p$. Then
we have that

$\log N\geq p^{2}\left(\frac{7}{3}-\frac{2.51}{t}-\frac{2.5\log
t}{t^{2}}-\frac{1.31}{t^{2}}-\frac{3.2\log t}{t^{3}}-\frac{3.2\log
t}{t^{3}}\right).$ (111)

Note that we have used OR($\frac{8}{3}$,$\frac{7}{3}$, $\\{3\\}$) rather than
our new bound since our main theorem is not better until $\omega\geq 34$. One
can derive a similar result, using the main theorem of this paper which will
be weaker when $N$ is divisible by a small prime $p$.

##  6 On the strength of restrictions about an odd perfect number

At this point, there are many different bounds on odd perfect numbers. These
include bounds on the size of the odd perfect number in terms of its number of
prime factors, bounds on the size of the largest prime factor, bounds on the
size of the smallest component and bounds on the size of $N$ itself. For a
given set of positive integers $A$, we will write $A(x)$ to be the number of
elements in $A$ which are at most $x$. Set $E$ to be the set of numbers of
Euler’s form for an odd perfect number. That is, $n\in E$ if $n=p^{a}m^{2}$
where $p$ is prime, $p\equiv a\equiv 1$ (mod 4), and $(p,m)=1$ Let $P$ be a
given property of a positive integer. We will write $E_{P}$ to be set of
elements of $E$ satisfying $P$. We will say that $P$ is a strong property if
the density of $E_{P}$ in $E$ is 0, that is

$\lim_{x\rightarrow\infty}\frac{E_{P}(x)}{E(x)}=0.$

We will similarly say that $P$ is a weak property if

$\lim_{x\rightarrow\infty}\frac{E_{P}(x)}{E(x)}=1.$

Note for example that for any constant $k$, all of the following are weak
properties:

  * •

“A number must be at least $k$”

  * •

“A number must have a prime factor at least $k$”

  * •

“A number must have a component at least $k$”

  * •

“A number must have at least $k$ distinct primes factors.”

  * •

“A number must have at least $k$ total prime factors.”

Note any finite set of weak properties cannot prove that no odd perfect
numbers exist.

However, Ochem and Rao’s inequality is in fact a strong property. Define
$OR_{\alpha,\beta}(n)$ to be the sentence
“$\Omega(n)\geq\alpha\omega(n)+\beta$.” It is a not difficult consequence of
Theorem 430 in [8] to show the following:

######  Theorem 59.

Let $\alpha$ and $\beta$ be real numbers. Assume that $\alpha>2$. Then
$E_{OR}$ is a strong property.

We will say that a property $P$ is _substantially stronger_ than property $Q$
if two conditions hold:

  1. 1.

Every element of $E$ which is satisfied by $Q$ is satisfied by $P$.

  2. 2.

The set $E_{P}$ has density zero in the set $E_{Q}$. That is,

$\lim_{x\rightarrow\infty}\frac{E_{P}(x)}{E_{Q}(x)}=1.$

We then conjecture that:

######  Conjecture 60.

Let $\alpha_{1},\alpha_{2}$, $\beta_{1},\beta_{2}$ be real numbers with
$\alpha_{1}>\alpha_{2}>2$. Then $OR_{\alpha_{1},\beta_{1}}$ is substantially
stronger than $OR_{\alpha_{2},\beta_{2}}$.

Of course, any result of the form “For any odd perfect $N$ $N$ must satisfy
$OR_{\alpha,\beta}$” cannot by itself resolve the fundamental open question,
but we suspect that the strength of Ochem and Rao’s result in the sense above
is a sign that this is a potentially fruitful direction for further research.
We note that something being a strong property does not always line up with
out intuition about what should be a “strong” property in a general sense. For
example, let $f(x)$ be a function which is increasing for sufficiently large
$x$ and satisfying that $\lim_{x\rightarrow\infty}f(x)=\infty$. Then it is not
hard to show that the property $P_{f}$ “For all $n$, $n$ has a prime factor
which is smaller than $f(n)$” is always a weak property. But if one could show
that an odd perfect number had to have prime factor always less than
$\log\log\log\log n$, that would certainly be noteworthy!

##  7 Future work and related problems

One major direction for improving these results is to prove there are no
triple threats. Proving there are no triple threats would result in
substantial tightening of both the bounds for the case of $3|N$ and for the
case of $3\not|N$. Another natural object of study in this context would be
what we call an $n$-obstruction.

Define an $n$-_obstruction_ to be a set of primes all greater than $3$
$a_{i},b_{i},c_{i}$ for $1\leq i\leq n$ and $p$ an odd prime, satisfying for
all $1\leq i\leq n$

  1. 1.

$\sigma(a_{i}^{2})=p\sigma(b_{i}^{2})\sigma(c_{i}^{2})$

  2. 2.

$\sigma(b_{i}^{2})$ and $\sigma(c_{i}^{2})$ prime.

  3. 3.

The $a_{i}$ are all distinct.

If we can show that a $4$-obstruction does not exist, possibly with some very
small modulo restrictions we will get a substantially tighter bound.
Similarly, if we can rule out $3$-obstructions or even a $2$-obstruction
(although we suspect that the last isn’t really doable). Note that at present
we can’t even show the following statement which looks like it should be
obviously true:

######  Conjecture 61.

There exists some $n$ such that no odd perfect number $N$, contains an
$n$-obstruction $a_{i},b_{i},c_{i}$ with $a_{i}^{2}||N$, $b_{i}^{2}||N$ and
$c_{i}^{2}||N$.

Of course, as we improve the linear term in the bounds, the general price paid
is that we are subtracting more in the constant term. Thus, in the original
Ochem and Rao paper, they had a constant of $-31/7$, and in the subsequent
paper we had as worst case constant $-39/8$. One of the original goals of
Ochem and Rao’s original inequality 1 was to assist in the proving of
inequality 2, and there is interest in proving inequalities of the form

$\Omega(N)\geq 2\omega(n)+C$ (112)

where $C$ is reasonably large. At present, the best such inequality is that by
Ochem and Rao where $C=51$.

Inequalities of that form require extensive computation, where one needs to
check many cases with branching in essentially the standard approach to heavy
computations to bound odd perfect numbers; however Ochem and Rao had as one of
their conditions to terminate a branch that Equation 1 forced Equation 2.
Obviously, that sort of termination will be more common when one has not just
a stronger linear term but a stronger constant term. Using the inequalities
from this paper to prove inequalities for the form of inequality 112 would be
easier with less negative constants. For specific small values of $\omega$ our
inequalities will already give slightly better bounds than used here, but
other approaches might improve the constants.

One might hope to use the results of Nielsen bounding the actual size of an
odd perfect number. We present here an approach that is too weak to be useful
by itself but might be productive with more work. We will restrict this
discussion under the assumption where that we have both $5|N$ and $11|N$ where
this approach is most likely to work. Assume further that we have $\omega=10$
which is the smallest possible value of $\omega$ not yet ruled out. Note that
if $5|\sigma(11^{f_{11}})$ or $11|\sigma(5^{f_{5}})$, then we can already
improve our constant term that way, so we will assume that neither of those
occurs. In that case we have

$(5^{f_{5}})^{2}(11^{f_{11}})^{2}<(5^{f_{5}})\sigma(5^{f_{5}})11^{f_{11}}\sigma(11^{f_{11}})\leq
N.$ (113)

Nielsen [12] has proved that if $N$ is an odd perfect number with $k$ distinct
prime factors and largest prime divisor $P$ then

$10^{12}P^{2}N<2^{(4^{k})}.$ (114)

Combining Equation 113 with Nielsen’s upper bound 114, as well as the fact
that the largest prime factor of an odd perfect number must be at least
$10^{8}$ by [6] we get that

$(5^{f_{5}})^{2}(11^{f_{11}})^{2}10^{28}<2^{(4^{10})}$ (115)

which when we take logarithms simplifies to

$(2\log_{2}5)f_{5}+(2\log_{2}11)f_{11}+28\log_{2}10<4^{10}$ (116)

which is is a linear inequality restricting $f_{5}$ and $f_{11}$ but it is
much too weak to give a useful restriction for improving the constant.

There appear to be four possible approaches to improving this inequality. The
first approach is that one could improve the size of the largest prime factor
of an odd perfect number; this is a project that should be undertaken in in
general since it has been about a decade since the last substantial
improvement on this has occurred; more recent algorithmic improvements and
computational power may make this a reasonable step. Unfortunately, it is
unlikely that such improvement by itself would substantially improve
Inequality 115 since the restriction involves the logarithm of the largest
prime divisor. The second approach is to improve the size of the largest prime
divisor, restricted to some specific range of $\omega$; it seems very likely
that with the additional assumption that $\omega=10$ or even something like
$\omega\leq 15$, that one can substantially improve on the lower bound for the
largest prime factor. The third possibility is to use Nielsen’s general
machinery which he used to prove Equation 114 to incorporate specific prime
powers. The fourth possibility is to improve the second inequality in Equation
113 by making the precise the intuition that there should be a large part of
$N$ which is not included in $\sigma(5^{f_{5}})\sigma(11^{f_{11}}).$ This last
looks to be the most promising. However, given how weak equation 115 is, it
will likely require multiple of these approaches for it to be at all
productive. Even if one improves it enough to be useful for small values of
$\omega$, it will still be likely too weak to be useful for even slightly
larger values of $\omega$. Luckily, all four of these approaches would be of
general interest to understanding odd perfect numbers. A slightly different
approach to Nielsen’s bound may also be valid. Again restricting to the
situation where $5|N$ and $11|N$, we have

$5^{f_{5}}11^{f_{11}}7^{2s}7^{4t}7^{6u}q^{e}<N$ (117)

and to then proceed as before. This inequality is also still too weak to be
directly useful by itself but may be combined with bounds on the size of $q$.
A major part of our improvement in the case of $3\not|N$ depended on a
specific coincidental factorization of a specific composition of cyclotomic
polynomials. Further understanding of such compositions may be relevant for
further understanding of odd perfect numbers. These questions about cyclotomic
polynomials may be of interest independent of anything involving perfect
numbers. We have a conjecture that essentially says that we cannot often get
so lucky that we often have such factorizations. In particular:

######  Conjecture 62.

Let $p$ and $q$ be distinct odd primes and let $\Phi_{p}(x)$ and $\Phi_{q}(x)$
be the $p$th and $q$th cyclotomic polynomials. Then at least one of
$\Phi_{p}(\Phi_{q}(x))$ or $\Phi_{q}(\Phi_{p}(x))$ is irreducible.

We also suspect that, in some suitable sense, such compositions being
reducible should occur on a set of density zero. In particular, call an
ordered pair of positive integers $(m,n)$ to be a _good_ if
$\Phi_{m}(\Phi_{n}(x))$ factors over the integers where $\Phi_{m}$ and
$\Phi_{n}$ are the $m$th and $n$th cyclotomic polynomials. Let $D(t)$ count
the number of good pairs with both $m\leq t$ and $n\leq t$. Then we strongly
suspect that:

######  Conjecture 63.

$\lim_{t\rightarrow\infty}\frac{D(t)}{t^{2}}=0.$

Moreover, we have the following stricter version: let $f(t)$ and $g(t)$ be
strictly increasing functions which go to infinity as $t$ goes to infinity,
and let $D_{f,g}(t)$ count the number of good pairs with $m\leq f(t)$ and
$n\leq g(t)$. Note that in particular $D(t)=D_{t,t}(t)$). Then

######  Conjecture 64.

For any such $f(t)$ and $g(t)$ we have

$\lim_{t\rightarrow\infty}\frac{D_{f,g}(t)}{f(t)g(t)}=0.$

We are uncertain if Conjecture 64 is true, but suspect that if it is true,
proving it would be very difficult. We can make corresponding versions of
Conjectures 63 and 64 that are restricted to cyclotomic polynomials arising
from primes. Define $\bar{D}(t)$ to be the same as $D(t)$ but counting only
the good pairs $(m,n)$ where $m$ and $n$ both prime. Define $\bar{D}_{f,g}(t)$
similarly. Then we expect that

######  Conjecture 65.

$\lim_{t\rightarrow\infty}\frac{\bar{D}(t)}{\pi(t)^{2}}=0.$

######  Conjecture 66.

For any such $f(t)$ and $g(t)$ we have

$\lim_{t\rightarrow\infty}\frac{\bar{D}_{f,g}(t)}{\pi(f(t))\pi(g(t))}=0.$

Note that similar questions have been asked and answered about general
polynomials. See in particular [22] and [18]. Ochem and Rao type results also
show that many divisors of a positive integer must have many repeated prime
factors. It is therefore of interest whether this sort of result can be used
to improve on results like [11] which rely heavily on inducting on the
divisors of an odd perfect number. One other obvious question is whether
anyone can replace the Ochem and Rao type results with a better than linear
inequality. The methods used in this paper do not seem to have any hope of
doing so, but it is plausible that sieve theoretic methods could result in
some similar type of restriction. One obvious question is how well we can
upper bound $\Omega(N)$ in terms of $\omega(N)$. Recall Nielsen’s result[11]
that if $N$ is an odd perfect number then

$N<2^{4^{\omega(N)}}.$ (118)

If $N$ is an odd perfect number then we trivially $3^{\Omega(N)}<N$, which
when combined with Equation 118 we obtain

$\Omega(N)<4^{\Omega(N)}\frac{\ln 2}{\ln 3}.$

Improving this bound directly in a non-trivial fashion seems worth exploring.
Nielsen also showed [11] that if $N$ is an odd perfect number, and we have
$P=\prod_{p|N}p$, then

$N<P^{2^{\omega(N)}},$ (119)

from which it follows that we have $a_{i}\leq 2^{\omega(N)-2}$ for at least
one of the $a_{i}$. It may be possible to use this fact to improve the Ochem-
Rao results further. We may also combine the Ochem and Rao type bounds to get
a straightforward upper bound for $N$ in terms of $\omega$. In particular, if
we know that $\Omega\geq\alpha\omega+\beta$ then we easily have from Equation
LABEL:Nieseln_bound that

$N<2^{4^{(\frac{\Omega-\beta}{\alpha})}}.$ (120)

Using our main theorem we have the result that if $(3,N)=1$ that

$N<2^{4^{\left(\frac{113\Omega+286}{302}\right)}}.$ (121)

It seems worth wondering if we can substantially improve upper bounds on $N$
in terms of $\Omega$ which are better than simply combining the Nielsen bound
with the best available Ochem and Rao type bound.

Ochem and Rao also used similar techniques in their proof that an odd perfect
number must have a component of size at least $10^{62}$.[13] In particular,
they first showed that any odd perfect number $N$ must either have a component
of size greater than $10^{62}$ or that $N$ cannot be divisible by any prime
less than $10^{8}$. They then concluded that an odd perfect number with all
components smaller than $10^{62}$ can only have primes raised to the first,
2nd, 4th or 6th powers. They then they obtained a set of linear inequalities
relating how many such primes there were and obtained a contradiction. It is
likely that the type tightened bounds in [23] and this paper can be used to
improve that type bound.

An additional area of interest may be to generalize the Ochem and Rao type
results beyond odd perfect numbers. Recall a number $N$ is said to be multiply
perfect if $Nk=\sigma(N)$ for some $k$, and we then say that $N$ is
$k$-perfect. Perfect numbers are then $2$-perfect. It is a long-standing
question if the only multiply perfect odd number is $1$. We suspect that the
the Ochem and Rao type results can be extended to odd multiply perfect numbers
where the constant term is allowed to be a function of $k$.

A different generalization of perfect numbers leads to Ore harmonic numbers.
Ore noted that if $N$ is a perfect number then one must have
$\sigma(N)|n\tau(n)$ where $\tau(n)$ is the number of positive divisors of
$n$. Ore called numbers satisfying $\sigma(N)|n\tau(n)$ harmonic numbers since
they are precisely the numbers where the harmonic mean of their positive
divisors is an integer. Note that there are multiply perfect numbers which are
not Ore harmonic numbers and there are Ore harmonic numbers which are not
multiply perfect numbers. Ore asked if all Ore harmonic numbers are odd. It
would be interesting to see if one can extend the Ochem and Rao type results
to Ore harmonic numbers. One can also generalize Ore’s harmonic numbers. We
will call $n$ a generalized harmonic number if $n$ satisfies
$\sigma(n)|n(\tau(n))g^{m}$ where $m$ is some integer and $g$ is the largest
odd divisor of $\tau(n)$. It again appears that all solutions here are odd,
although as far as we are aware, this generalization has not been investigated
in the literature. It would be interesting to see if Ochem and Rao type bounds
can be extended to these generalized harmonic numbers.

Another interesting direction is rather than generalize instead to narrow the
situation. Colton [2] has shown that no perfect number (whether even or odd)
satisfies $\tau(n)|n$. However, the set of positive integers which satisfy
$\tau(n)|n$ have density zero.[9] In contrast, the set of numbers $n$ where
$\tau(n)|\sigma(n)$ has density 1. [1] It is not hard to show that the only
even perfect number $n$ satisfying $\tau(n)|\sigma(n)$ is $n=6$. One might ask
if we can say anything interesting about odd perfect numbers $N$ satisfying
$\tau(N)|\sigma(N)$. In particular, it is likely that Ochem-Rao type results
can be substantially improved if one is restricted to this set.

Acknowledgements: The author is grateful to detailed feedback from Pascal
Ochem which substanially improved the presentation as well as the strength of
the results. Aaron Silberstein first alerted the author to the fact that a
composition of cyclotomic polynomials can be reducible. Douglas McNeil pointed
out that an earlier version of Lemma 7 was incorrect, and also pointed out
that a previous version’s conjecture was hopelessly false.

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