# The Mobius Function and Congruent Numbers

Roy Burson

###### Abstract

This work provides a complete characterization of congruent numbers in terms
of Pythagorean triples. Specifically, we show that every congruent number can
be written as

$\frac{nm\left(m-n\right)\left(m+n\right)}{\sigma^{2}}$

were as

$\sigma|\rho\biggl{(}\left(m-n\right)\left(m+n\right)\biggr{)},\indent\text{or}\indent\sigma|\rho(nm)$

were $\rho(\alpha)$ denotes the non-square free part of its argument $\alpha$.
As a consequence, in order to find congruent numbers it suffices to devise a
condition so that the equality $\mu(m-n)+1=\gcd(m,n)$ or
$\mu(m+n)+1=\gcd(m,n)$ holds, were $\mu$ is the Mobius function.

###### keywords:

Congruent Numbers , The Mobius Function

††journal: Number Theory

##  1 Introduction

According to [9] the set of congruent numbers is one of the oldest sets of
numbers on record first appearing in the Arab Manuscripts. The most recent
results concerning congruent numbers stems from the properties of elliptic
curves [5], which is connected to the Birch and Swinnerton-Dyer Conjecture
discussed by [6]. Assuming the validity of the Birch and Swinnerton-Dyer
Conjecture conjecture [7] derived a criteria to determine if a given number
$n$ is congruent. Despite of this result, the congruent number problem has yet
to be solved. The result in [7] has brought a wide spread attention to the
properties of elliptic curves.
Due to this wide spread attention surrounding the group properties of elliptic
curves no new elementary results have been seen in the recent years. Instead
of developing any new ideas most authors have pushed there studies toward a
solution of the Birch and Swinnerton-Dyer Conjecture by studying the group
properties of Elliptic curves and attempting to address the rank, which is not
the case in this paper. In this work we address the congruent problem by
developing some new results, which have not been seen in other literature. The
results obtained here are elementary as they are independent of any previous
results surrounding congruent numbers except for Euclid’s theorem that
establishes a parametric description of Improper-Congruent numbers (which we
discuss later) and [8]’s quick discuss on how to find square free congruent
numbers. [8] actually suggest that using the Pythagorean theorem is a bad idea
to approach the congruent number problem, which this work proves wrong by
showing that congruent numbers are specific divisors of Pythagorean triples.
This paper serves to provide the complete parametric description of all
congruent numbers using Pythagorean triples. Specifically, this work shows
that all congruent numbers are either Pythagorean triples or they are the non-
square free divisors of Pythagorean triples. Hence, in order to find congruent
number this work shows it suffices to examine the non-square free divisors of
the Pythagorean triples. Thus providing a direct link between Pythagorean
triples and all congruent numbers. Although this work provided the complete
analytical description of congruent numbers non of these results are connected
to Elliptic curves as it would overcompensate the procedure presented herein
as it is long enough.
The main objective of this work is too separate the congruent numbers into
three different categories: Improper, Semi-Proper, and Proper. In each case we
gather some results independently and then bring them all together towards the
end of the work. Improper congruent numbers are those congruent numbers which
represent the area of a triangle which has both legs strictly in $\mathbb{Z}$.
Semi-Proper congruent numbers are those congruent numbers which represent the
area of a triangle with one leg in $\mathbb{Z}$ and the other strictly in
$\mathbb{Q}$. Proper-Congruent numbers are those congruent numbers which has
all sides in $\mathbb{Q}$ and strictly not in $\mathbb{Z}$.
In section 2 we develop some preliminary results that are needed throughout
the work. Moreover the ideas in this section have shaped or formed the rest of
the work. In this section we discuss Euclid’s famous theorem (Theorem $\S$
2.1) regarding the parametric description of Pythagorean triples and show how
it can be used to develop some results concerning proper congruent numbers.
The main result in this section is provided by Theorem $\S$ 2.2, which
provides a condition to find the Proper-Congruent numbers. This result
combined with Proposition $\S$ 2.2 shows that there is a correspondence
between the proper congruent numbers and the divisors of the Pythagorean
triple, which is discussed more thoroughly in Section 7.
Section 3 shows how one can use the Mobius function to generate some congruent
numbers. This section established some preliminaries which help establish a
connection between the proper congruent numbers and the altered Fermat
equation $ax^{4}+by^{4}=z^{2}$ discusses in the latter section.

##  2 Primitive Triples and Proper Congruent Numbers

######  Definition 2.1.

A Pythagorean triple is a set of integers $\\{a,b,c\\}$ so that
$a^{2}+b^{2}=c^{2}$ were as $a,b,c\in\mathbb{Z}$ (for, example $(3,4,5)$). If
$\gcd(a,b,c)=1$ we say that $\\{a,b,c\\}$ is a Primitive triple.

######  Theorem 2.1.

(Euclid’s Elements [3]). All primitive Pythagorean triples $(a,b,c)$ have the
representation

$\displaystyle a$ $\displaystyle=m^{2}-n^{2}$ $\displaystyle b$
$\displaystyle=2nm$ $\displaystyle c$ $\displaystyle=m^{2}+n^{2}$

were $\gcd(n,m)=1$ and $m>n$ and $m\not\equiv n(\bmod{2})$.

######  Definition 2.2.

A congruent number is a number $n\in\mathbb{N}$ that is the area of some right
triangle $\triangle ABC$ with rational sides
$(\frac{a}{b},\frac{c}{d},\frac{e}{f})$.

######  Definition 2.3.

A Proper-Congruent number is a number $n\in\mathbb{N}$ that represents the
area of some right triangle $\triangle ABC$ with rational sides
$(\frac{a}{b},\frac{c}{d},\frac{e}{f})$ and the additional property that
neither $b$ nor $d$ can be one. That is $b\neq 1\neq d$. So both sides lengths
of $\triangle ABC$ are strictly in $\mathbb{Q}$ and not $\mathbb{Z}$.

######  Proposition 2.1.

If $n$ is a Proper-Congruent number with $\triangle ABC$ constructed by the
triple $(\frac{a}{b},\frac{c}{d},\frac{e}{f})$, then $f=bd$.

###### Proof.

Suppose $n$ is a proper congruent number with
$1=\gcd(a,b)=\gcd(c,b)=\gcd(e,f)$. Since $n$ is congruent it represents the
area of the triangle $\triangle ABC$ which has area
$n=\frac{1}{2}(\frac{a}{b})(\frac{c}{d})=\frac{ac}{bd}$. Now $n$ is an integer
so $bd|ac$. Since $n$ is proper we know that $b$ does not divide $c$ and $d$
does not divide $a$. Therefore $d|a$ and $b|c$. Let $i$ denote the largest
power of $b$ so that $b^{i}|c$ but $b^{i+1}\not|c$. Similarly, let $j$ denote
the largest power of $d$ so that $d^{j}|a$ but $d^{j+1}\not|a$. Write
$a=d^{j}q_{2}$ and $c=b^{i}q_{1}$ for some quotients $q_{1}$ and $q_{2}$, then
by the hypothesis it follows that

$\displaystyle(\frac{a}{b})^{2}+(\frac{c}{d})^{2}$
$\displaystyle=(\frac{e}{f})^{2}$ $\displaystyle\Downarrow$
$\displaystyle(\frac{d^{j}q_{2}}{b})^{2}+(\frac{b^{i}q_{1}}{d})^{2}$
$\displaystyle=(\frac{e}{f})^{2}$ $\displaystyle\Downarrow$
$\displaystyle\frac{d^{2j}q_{2}^{2}}{b^{2}}+\frac{b^{2i}q_{1}^{2}}{d^{2}}$
$\displaystyle=\frac{e^{2}}{f^{2}}$ $\displaystyle\Downarrow$
$\displaystyle\frac{d^{2j+1}q_{2}^{2}+b^{2i+2}q_{1}^{2}}{b^{2}d^{2}}$
$\displaystyle=\frac{e^{2}}{f^{2}}$

Now suppose that the numerator and the denominator of the last equality are
not equal. Then it must follow that

$b^{2}d^{2}|\left(d^{2j+1}q_{2}^{2}+b^{2i+2}q_{1}^{2}\right)$

but $bd|b^{2}d^{2}$ so then

$bd|b^{2}d^{2}|\left(d^{2j+1}q_{2}^{2}+b^{2i+2}q_{1}^{2}\right)$

Thus product $bd$ divides the numerator. This implies $b$ and $d$ divide the
numerator individually. This clearly cannot happen because $b$ divides the
numerator if and only if $b|q_{1}^{2}$, but $a=d^{j}q_{1}$ and
$1=\gcd(a,b)=\gcd(b,d^{j}q_{1})$ so $b$ cannot divide $q_{1}$. The same
argument holds for the integer $d$. Thus the numerator and the denominator are
equal, which implies that is $f^{2}=b^{2}d^{2}=(bd)^{2}$. Hence, $f=bd$ as
desired. ∎

######  Proposition 2.2.

If $n$ is a proper congruent number with $\triangle ABC$ constructed by the
triple $(\frac{a}{b},\frac{c}{d},\frac{e}{f})$, then
$e^{2}=(da)^{2}+(bc)^{2}$.

###### Proof.

Assume that $n$ is a congruent number with the right triangle $\triangle ABC$
constructed by the triple $(\frac{a}{b},\frac{b}{c},\frac{e}{f})$. By
proposition $\S$ 2.1 it follows that $f=bd$. Therefore,

$\displaystyle(\frac{e}{f})^{2}$
$\displaystyle=(\frac{a}{b})^{2}+(\frac{c}{d})^{2}$ $\displaystyle\Downarrow$
$\displaystyle(\frac{e}{bd})^{2}$
$\displaystyle=(\frac{a}{b})^{2}+(\frac{c}{d})^{2}$ $\displaystyle\Downarrow$
$\displaystyle e^{2}$
$\displaystyle=(bd)^{2}\left((\frac{a}{b})^{2}+(\frac{c}{d})^{2}\right)$
$\displaystyle\Downarrow$ $\displaystyle e^{2}$
$\displaystyle=(bd)^{2}(\frac{a}{b})^{2}+(bd)^{2}(\frac{c}{d})^{2}$
$\displaystyle\Downarrow$ $\displaystyle e^{2}$
$\displaystyle=(da)^{2}+(bc)^{2}$

∎

######  Remark 2.1.

One is tempted to believe that Proposition $\S$ 2.1 and $\S$ 2.2 follow
directly from the Pythagorean theorem. However, this is not true. For example,
if we have a right triangle $\triangle ABC$ with sides
$(\frac{a}{b},\frac{c}{d},\frac{e}{f})$ then the Pythagorean theorem says that
$(\frac{e}{f})^{2}=(\frac{a}{b})^{2}+(\frac{c}{d})^{2}=\frac{(ad)^{2}+(bc)^{2}}{(bd)^{2}}$.
One may now be attempted to assume that $e=\sqrt{(ad)^{2}+(bc)^{2}}$ and
$f=bd$ but we cannot jump directly to this conclusion because this is not a
general result that holds. For example, if $\frac{x}{y}=\frac{k}{l}$ then
$x=k$ and $y=l$ if and only if $\gcd(x,y)=1=\gcd(k,l)$, and here we do not
know the greatest common factor so we may not make this conclusion. However,
it turns out to be true as we will demonstrate below.

######  Proposition 2.3.

If $(a,b,c)$ is a primitive Pythagorean triple then, $\gcd(a,b)=1$.

###### Proof.

Suppose $(a,b,c)$ is a primitive triple with $\gcd(a,b)\neq 1$. Then set
$d=\gcd(a,b)>1$. By the Euclidean division algorithm there exist $k_{1}$,
$k_{2}$ so that $a=dk_{1}$ and $b=dk_{2}$. Then

$\displaystyle c^{2}$ $\displaystyle=a^{2}+b^{2}$ $\displaystyle\Downarrow$
$\displaystyle c^{2}$ $\displaystyle=(dk_{1})^{2}+(dk_{2})^{2}$
$\displaystyle\Downarrow$ $\displaystyle c^{2}$
$\displaystyle=d^{2}(k_{1}^{2}+k_{2}^{2})$ $\displaystyle\Downarrow$
$\displaystyle(\frac{c}{d})^{2}$ $\displaystyle=k_{1}^{2}+k_{2}^{2}$

Since $k_{1}$ and $k_{2}$ are integers it must follow that $d|c$. But this
means $\gcd(a,b)|c$. Since $\gcd(a,b)|a$ and $\gcd(a,b)|b$ it follows that
$d=\gcd(a,b)$ is a divisor or $a$, $b$, and $c$. So
$\gcd(a,b,c)\geq\gcd(a,b)>1$. So then $\gcd(a,b,c)>1$ which is to say that
$(a,b,c)$ is not primitive. This is a contradiction. ∎

######  Theorem 2.2.

Let $a,b,c,d,e$ be integers that satisfy the following:
(i) $e^{2}=(da)^{2}+(bc)^{2}$
(ii) $d|a$, $b|c$, $\gcd(a,b)=1=\gcd(c,d)$, and $b\neq 1\neq d$.
(iii) $\\{a,c\\}$ contains an even integer, and this even number has more
factors of 2 than $b$ or $d$.
Then the number $\frac{1}{2}(\frac{a}{b})(\frac{c}{d})$ is a proper congruent.

###### Proof.

Assume $(i)-(iii)$ are valid. Set $f=bd$ and observe that

$\displaystyle e^{2}$ $\displaystyle=(da)^{2}+(bc)^{2}$
$\displaystyle\Downarrow$ $\displaystyle\frac{e^{2}}{f^{2}}$
$\displaystyle=\frac{(da)^{2}+(bc)^{2}}{f^{2}}$ $\displaystyle\Downarrow$
$\displaystyle\frac{e^{2}}{f^{2}}$
$\displaystyle=\frac{d^{2}a^{2}}{f^{2}}+\frac{b^{2}c^{2}}{f^{2}}$
$\displaystyle\Downarrow$ $\displaystyle\frac{e^{2}}{f^{2}}$
$\displaystyle=\frac{a^{2}}{b^{2}}+\frac{c^{2}}{d^{2}}$
$\displaystyle\Downarrow$ $\displaystyle(\frac{e}{f})^{2}$
$\displaystyle=(\frac{a}{b})^{2}+(\frac{c}{d})^{2}$

This is a rational right triangle with area
$\frac{1}{2}(\frac{a}{b})(\frac{c}{d})$. This is a whole integer since $d|a$
and $b|c$ and by condition (iii) the number $a$ or $c$ contains at least one
more factor of $2$ its divisor $b$ or $d$, which also satisfy the condition
$b\neq 1\neq d$. We only need to very that $\gcd(e,f)=1=\gcd(a,b)=\gcd(c,d)$.
Condition (ii) gives that $1=\gcd(a,b)=\gcd(c,d)$ so we only need to show
$\gcd(e,f)=1$. By condition (i) we can see that $bd$ does not divide $e^{2}=$
since $e^{2}=(da)^{2}+(bc)^{2}$. So $bd$ cannot divide $e$, which means
$\gcd(f,e)=\gcd(bd,e)=1$. ∎

######  Example 2.1.

Let us try to find a Pythagorean triple that satisfies the conditions of the
hypothesis in Theorem $\S$ 2.1 and $\S$ 2.2. According to [1], around 1220
Fibonacci was challenged by Johannes Palermo to find a right triangle whose
are is equal to $5$, and Fibonacci found
$\\{\frac{3}{2},\frac{20}{3},\frac{41}{6}\\}$. Nobody actually knows how he
came up with this triangle, however, he stated that if $n$ is perfect square
then $n$ is not a congruent number. The proof had to wait until (1601-1665)
when Pierre de Fermat finally solved the problem.
Now let us attempt to find how Fibonacci found this triangle using Theorem
$\S$ 2.1 and $\S$ 2.2. By Theorem $\S$ 2.1 it follows that all the primitive
Pythagorean triples $(a,b,c)$ have the representation

$\displaystyle a$ $\displaystyle=m^{2}-n^{2}$ $\displaystyle b$
$\displaystyle=2nm$ $\displaystyle c$ $\displaystyle=m^{2}+n^{2}$

were $\gcd(n,m)=1$ and $m>n$ and $m\not\equiv n(\bmod{2})$. Choose $m=5,n=4$
then $\gcd(5,4)=1$, $5\neq(4{\bmod{2}})$, and $5>4$ so these integers qualify
to an integer Pythagorean triple. That is $a=(m-n)(m+n)=1\cdot 9=9$, $b=2\cdot
m\cdot n=2\cdot 3\cdot 2=12$. Both $a$ and $b$ are square free so we can break
them up by factoring the largest non square free part of both $a$ and $b$ as
$a=3\cdot 3=9$ and $b=2\cdot 20$. Write $c^{2}=(9)^{2}+(40)^{2}=(3\cdot
3)^{2}+(2\cdot 20)^{2}$, then by direct application of Theorem $\S$ 2.2 it
follows that the integer $n=\frac{1}{2}(\frac{3}{2})(\frac{20}{3})=5$ is a
congruent number generated by the triple
$(\frac{3}{2},\frac{20}{3},\frac{41}{6})$. This is precisely the exact triple
that Fibonacci discovered in 1220 in response to his challenger.

######  Definition 2.4.

We say a number $n$ is square free if $p^{2}\not|n$ for all $p|n$. A number is
not square free if there is a divisor $p\neq 1$ of $n$ so that $p^{2}|n$.

######  Lemma 2.1.

Assume that $m,n\in\mathbb{Z}$. If $\gcd(m,n)=1$, then $\gcd(m-n,m+n)=1$ or
$\gcd(m-n,m+n)=2$ .

###### Proof.

Assume that $m,n\in\mathbb{Z}$ were as $\gcd(m,n)=1$. Let $d$ be a common
positive divisor of $m+n$ and $m-n$. Then $d|m+n$ and $d|m-n$. Hence,
$d|(m+n)+(m-n)=2m$. Similarly, $d|(m+n)-(m-n)=2n$. So we have that $d|2n$ and
$d|2m$. Since $\gcd(m,n)=1$ $d$ must divide $2$. So then $d=1$ or $d=2$. ∎

##  3 Generation

######  Definition 3.1.

Let $n\in\mathbb{N}$. The Mobius function is the map

$\mu(n)=\begin{cases}1&\textit{if $n$ is square free}\\\ 0&\textit{if $n$ is
not square free}\\\ \end{cases}$

######  Conjecture 3.1.

If $n\in 2^{\mathbb{N}}$ then $n$ is not congruent.

######  Proposition 3.1.

Let $p$ be a prime number. If there exist $(k,l)$ so that
$k^{2}-64p^{2}=l^{2}$ were as $l>1$ then $p$ is a congruent number.

###### Proof.

Suppose $p$ is a prime number and suppose that there exist $(k,l)$ so that
$k^{2}-64p^{2}=l^{2}$. Write $k^{2}=l\cdot l+(2\cdot(4p))^{2}$. Define
$d=l=a$,$b=2$, and $c=4p$. Since $\gcd(l,8p)=1$ it follows that
$\gcd(d,4p)=1$. Since $d=l>1$ we see that $d$ does not divide $p$ (otherwise
we would have a contradiction). Thus $\gcd(a,b)=1=\gcd(c,d)$. Now by direct
application of Theorem $\S$ 2.2 it follows that the number
$n=\frac{1}{2}(\frac{a}{d})(\frac{4p}{2})$ is congruent. But since $a=d$ then
we trivially have $n=p$. Hence, we have shown that $p$ is congruent. ∎

######  Proposition 3.2.

If $|m-n|=1$, then $\gcd(m-n,m+n)=1$

###### Proof.

Let $|m-n|=1$ so that $m=n+1$ or $n=m+1$. In the first case we have
$\gcd(m-n,m+n)=\gcd((n+1)-n,(m+1)+m)=\gcd(1,2m+1)=1$. Similarly, in the second
case, if $n=m+1$ then,
$\gcd(m-n,m+n)=\gcd(m-(m+1),n+(n-1))=\gcd(1,2n-1)=1.(1,2n+1)$. These are the
only possibilities so we are done. ∎

######  Corollary 3.1.

If $\gcd(m-n,m+n)=2$, then $|m-n|\neq 1$

###### Proof.

This is the contropositive of proposition $\S$ 3.2. By Lemma $\S$ 2.1 if
$\gcd(m-n,m+n)\neq 1$ it must be that $\gcd(m-n,m+n)=2$. Since this is the
only other possibility. ∎

######  Proposition 3.3.

If $\gcd(m,n)=1$ and $m\not\equiv n(\bmod{2})$ then $\gcd(m-n,m+n)=1$.

###### Proof.

Assume $\gcd(m,n)=1$ and $m\not\equiv n(\bmod{2})$. By lemma $\S$ 2.1 it
follows that $\gcd(m,n)=1$ or $\gcd(m,n)=2$. Since $m\not\equiv n(\bmod{2})$
it follows that $m-n$ and $m+n$ are both odd. Hence, $2$ cannot divide neither
of these numbers. This leaves the only conclusion that $\gcd(m,n)=1$. ∎

######  Remark 3.1.

Theorem $\S$ 2.2 illustrates that there is some type of correspondence between
non-square free Pythagorean triples and congruent numbers (not necessarily a
bijection). In this sense, one can generate all some congruent numbers by
finding all non-square free Pythagorean triples. Later in definition $\S$ 2.1
we classify these types of congruent as Proper Congruent Numbers. Recall
Euclid’s Theorem 2.1 asserts that all Pythagorean triples $\\{a,b,c\\}$ have
the form

$\displaystyle a$ $\displaystyle=m^{2}-n^{2}$ $\displaystyle b$
$\displaystyle=2nm$ $\displaystyle c$ $\displaystyle=m^{2}+n^{2}$

were $\gcd(n,m)=1$ and $m>n$ and $m\not\equiv n(\bmod{2})$. Since $m\not\equiv
n(\bmod{2})$ it follows that $4|b$ and hence $\mu(b)=0$. This indicates that
one needs only to find when $a=m^{2}-n^{2}$ is not square free. But we now
$a=m^{2}-n^{2}=(m-n)(m+n)$ so that $\mu(a)=0$ whenever $\mu(m-n)=0$,
$\mu(m+n)=0$, or $\mu(m^{2}-n^{2})=0$. In the next proposition we will show
that $\mu(m^{2}-n^{2})=0$ can not happen,which leaves us to find when $m-n$ or
$m+n$ is not square free.

######  Example 3.1.

Consider $m=2^{2}\cdot 5^{2}\cdot 7^{3}=34300$ and $n=11^{2}\cdot
13^{2}=20449$. Then set $\beta=2mn=(2^{3}\cdot 5^{2}\cdot 7^{3})(11^{2}\cdot
13^{2})$ and compute $\alpha=m^{2}-n^{2}=(m-n)(m+n)=(13851)(54749)=(3^{6}\cdot
19)(53\cdot 1033)$. Clearly, $\mu(\beta)=0$ since $\mu(n)=0$ and $\mu(m)=0$,
and $\mu(\alpha)=0$ since $3^{6}|\alpha$. By Euclid’s Theorem $\S$ 2.1 we now
$(\alpha,\beta,\gamma)$ is a primitive Pythagorean triple. Henceforth, using
Theorems $\S$ 2.2 we can find a congruent number that corresponds to this
triple (Note! the theorem does not state that the congruent number is unique.
This is because on some occasions we may form more than one congruent number
for a given primitive triple $\\{\alpha,\beta,\gamma\\}$ (see table 4.1).
To find a corresponding congruent number we may take any divisors $a,d|\alpha$
and $b,c|\beta$ so that $(m^{2}+n^{2})^{2}=(da)^{2}+(bc)^{2}$ were $d|a$,
$b|c$. We can do this by rearranging the factorization of $\alpha$ and
$\beta$. For example, set $d=3$, $a=3^{5}\cdot 19$, $b=2\cdot 5\cdot 7$, and
$c=2^{2}\cdot 5\cdot 7^{2}\cdot 11^{2}\cdot 13^{2}$. Allow $e=m^{2}+n^{2}$
then $e^{2}=(da)^{2}+(bc)^{2}$ were $d|a$, $b|c$ and $d\neq 1\neq b$.
Corollary to Theorem $\S$ 2.2 it follows that the number
$n=\frac{1}{2}(\frac{a}{b})(\frac{c}{d})=\frac{1}{2}(\frac{3^{5}\cdot
19}{2\cdot 5\cdot 7})(\frac{2^{2}\cdot 5\cdot 7^{2}\cdot 11^{2}\cdot
13^{2}}{3})=2\cdot 3^{4}\cdot 7\cdot 11^{2}\cdot 13^{2}=23189166$ is
congruent. We could have also chosen $d=3$, $a=3^{5}\cdot 19$, $b=2$, and
$c=2^{2}\cdot 5^{2}\cdot 7^{3}\cdot 11^{2}\cdot 13^{2}$, so that we would have
found $n=\frac{1}{2}(\frac{a}{b})(\frac{c}{d})=\frac{1}{2}(\frac{3^{5}\cdot
19}{2})(\frac{2^{2}\cdot 5^{2}\cdot 7^{3}\cdot 11^{2}\cdot
13^{2}}{3})=1092566475$ instead of $23189166$ as above. To find all congruent
numbers that correspond to this specific Pythagorean triple we can list all
the different factorization and compute each different congruent number. The
results of each possible triple is illustrated in Table 3.1 below.

$n$ | $\frac{1}{2}(\frac{a}{b})(\frac{c}{d})$
---|---
$1943020219000$ | $\frac{1}{2}(\frac{3^{6}\cdot 19}{1})(\frac{2^{3}\cdot 5^{2}\cdot 7^{3}\cdot 11^{2}\cdot 13^{2}}{1})$
$485755054800$ | $\frac{1}{2}(\frac{3^{6}\cdot 19}{2})(\frac{2^{2}\cdot 5^{2}\cdot 7^{3}\cdot 11^{2}\cdot 13^{2}}{1})$
$215891135500$ | $\frac{1}{2}(\frac{3^{5}\cdot 19}{1})(\frac{2^{2}\cdot 5^{2}\cdot 7^{3}\cdot 11^{2}\cdot 13^{2}}{3})$
$53972783870$ | $\frac{1}{2}(\frac{3^{5}\cdot 19}{2})(\frac{2^{2}\cdot 5^{2}\cdot 7^{3}\cdot 11^{2}\cdot 13^{2}}{3})$
$5996975985$ | $\frac{1}{2}(\frac{3^{4}\cdot 19}{2})(\frac{2^{2}\cdot 5^{2}\cdot 7^{3}\cdot 11^{2}\cdot 13^{2}}{3^{2}})$
$122387265$ | $\frac{1}{2}(\frac{3^{3}\cdot 19}{2})(\frac{2^{2}\cdot 5^{2}\cdot 7^{3}\cdot 11^{2}\cdot 13^{2}}{3^{3}})$
$1011465$ | $\frac{1}{2}(\frac{3^{3}\cdot 19}{2\cdot 5})(\frac{2^{2}\cdot 5\cdot 7^{3}\cdot 11^{2}\cdot 13^{2}}{3^{3}})$
$5985$ | $\frac{1}{2}(\frac{3^{3}\cdot 19}{2\cdot 5\cdot 7})(\frac{2^{2}\cdot 5\cdot 7^{2}\cdot 11^{2}\cdot 13^{2}}{3^{3}})$
$8645$ | $\frac{1}{2}(\frac{3^{3}\cdot 19}{2\cdot 5\cdot 7\cdot 11})(\frac{2^{2}\cdot 5\cdot 7^{2}\cdot 11\cdot 13^{2}}{3^{3}})$
$665$ | $\frac{1}{2}(\frac{3^{3}\cdot 19}{2\cdot 5\cdot 7\cdot 11\cdot 13})(\frac{2^{2}\cdot 5\cdot 7^{2}\cdot 11\cdot 13}{3^{3}})$
Table 3.1. List of different congruent numbers that correspond to the triple
$(758328399,200400200,615222200900000000)$

To see why this is useful one has to notice that with a single Pythagorean
triple we were able to find $10$ congruent numbers shuffling the factorization
of each term in $\frac{1}{2}(\frac{a}{b})(\frac{c}{d})$. Ideally, one would
like to find large Pythagorean triples $(\alpha,\beta,\gamma)$ were
$\mu(\alpha)=0=\mu(b)$ and the number of different factors of $a$ and $b$ is
large. This is explored later in section 6 and 7.

######  Definition 3.2.

Suppose that $n\in\mathbb{N}$ and $a|n$ were $a\neq 1$. The mapping
$\varrho:\mathbb{N}\rightarrow\mathbb{N}$ defined by the rule
$n\mapsto\max\\{d:d|n,~{}n=d^{2}a,~{}a\neq 1\\}$ is called the Square Part of
$n$. It is the non-square free part of an integer. If $n$ is square free then
we decree that $\varrho(n)=1$.

######  Lemma 3.1.

let $p$ be a prime number then $\varrho(p^{i})=p^{\lfloor\frac{i}{2}\rfloor}$
for each $i\geq 2\in\mathbb{N}$.

###### Proof.

let $p$ be a prime number and consider $y=p^{i}$ for $i\geq 2\in\mathbb{N}$.
Now by definition $\varrho(p^{i})=\max\\{d:d|p^{i},~{}p^{i}=d^{2}a,~{}a\neq
1\\}$. But any divisor of $p$ is either $1$ or $p$. So the maximum divisor $d$
that fits this property must by at least $p$ if $i\geq 2$. Note that
$\left(p^{\lfloor\frac{i}{2}\rfloor}\right)\left(p^{\lfloor\frac{i}{2}\rfloor}\right)=p^{\lfloor\frac{i}{2}\rfloor+\lfloor\frac{i}{2}\rfloor}=p^{\lfloor\frac{2i}{2}\rfloor}=p^{i}$
and $\left(p^{\lfloor\frac{i}{2}\rfloor}\right)$ is the largest integer that
divides $\left(p^{\lfloor\frac{i}{2}\rfloor}\right)$. Hence,
$\varrho(p^{i})=p^{\lfloor\frac{i}{2}\rfloor}$ and we are done. ∎

######  Lemma 3.2.

The Square Part of $n$ is multiplicative over relatively prime numbers. That
is if $a,b\in\mathbb{N}$ so that $\gcd(a,b)=1$ then $\varrho(a\cdot
b)=\varrho(a)\cdot\varrho(b)$.

###### Proof.

Let $\gcd(a,b)=1$. If $\varrho(a)=1$ or $\varrho(b)=1$ then the result is
trivial. So assume that $\varrho(a)\neq 1$ and $\varrho(b)\neq 1$. This being
the case we can write $a=A^{2}k$ and $b=B^{2}l$ were $\varrho(a)=A^{2}$ and
$\varrho(b)=B^{2}$. Then $ab=A^{2}B^{2}kl=(AB)^{2}(kl)$. Moreover, since
$A^{2}|k$ and $B^{2}|l$ it follows that $(AB)^{2}|(kl)$. Since $\gcd(a,b)=1$
it must follow that $\gcd(AB,kl)=1$. Thus the square part of $ab$ is
$(AB)^{2}$ and hence
$\varrho(a)\cdot\varrho(b)=A^{2}B^{2}=(AB)^{2}=\varrho(ab)$ as desired. ∎

######  Proposition 3.4.

Let $n\in\mathbb{N}$ and write $n$ in conical form
$n=\prod_{i=1}^{k}{p^{e_{i}}_{i}}$ for primes $p_{1},p_{2},\cdots,p_{k}$ and
integers $e_{1},e_{2},\cdots,e_{k}\in\mathbb{N}$. Then

$\varrho(n)=\prod_{i=1}^{k}{p^{\lfloor\frac{e_{i}}{2}\rfloor}_{i}}$

###### Proof.

This proof utilities the multiplicative property of the $\varrho$. Write
$n=\prod_{i=1}^{k}{p_{i}^{e_{i}}}$ it its conical factorization. Since $p_{i}$
is prime for each $i\in\\{1,2,\cdots,k\\}$ it follows that
$\gcd(p_{i}^{e_{i}},p_{j}^{e_{j}})=1$ for any such $i$, $j$, $e_{i}$, and
$e_{j}$. Hence,

$\displaystyle\varrho(n)$ $\displaystyle=\varrho(\prod_{i=1}^{k}{p^{e_{i}}})$
$\displaystyle=\prod_{i=1}^{k}{\varrho(p^{e_{i}})}$
$\displaystyle=\prod_{i=1}^{k}{p^{\lfloor\frac{e_{i}}{2}\rfloor}}$

were the first equality follows by substitution of $n$ in its conical form,
the second equality holds by $\S$ 3.2 because each factor is relatively prime,
and the third holds by lemma $\S$ 3.1. ∎

##  4 Proper Congruent Numbers and the Altered Fermat Equation
$ax^{4}+by^{4}=z^{2}$

Fermat was the first to prove that the equation $x^{4}+y^{4}=z^{2}$ has no-non
trivial solution $(x,y,z)\in\mathbb{Z}^{3}$. In this section, I illustrate how
proper congruent numbers are related to the diophantine solutions of the
equation $ax^{4}+by^{4}=z^{2}$ for suitable integers $a$ and $b$.
Specifically, I show that a whole integer $n\in\mathbb{N}$ being congruent
relies heavily on the solutions of $ax^{4}+by^{4}=z^{2}$ were $\gcd(a,b)=1$,
and $\mu(a)=0=\mu(b)$ or simply just one of them, that is $\mu(a)=0$ or
$\mu(b)=0$. Furthermore, it is shown that $ax^{4}+by^{4}=z^{2}$ has a solution
if and only if $a+b=z^{2}$ which occurs if and only if there is a solution
$c\in\mathbb{N}$ so that $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ were
$\gcd(a,b,c)$.

######  Proposition 4.1.

Each proper congruent number gives rise to a solution of the equation
$ax^{4}+by^{4}=z^{2}$ were $\mu(a)=0=\mu(b)$ and $\gcd(a,b)=1=\gcd(c,d)$.

###### Proof.

First assume that $n$ is a proper congruent number. By definition there exist
$(\frac{a}{b},\frac{c}{d},\frac{e}{f})$ so that
$n=\frac{1}{2}(\frac{a}{b})(\frac{c}{d})$ were as $(da)^{2}+(bc)^{2}=e^{2}$
with the condition that $d|a$,$b|c$ and $d\neq 1\neq b$. This being the case
we may write $a=dk$ and $c=bl$ for some $(k,l)$ so that
$(da)^{2}+(bc)^{2}=e^{2}$ becomes $(d^{2}k)^{2}+(b^{2}l)^{2}=e^{2}$. This is a
solution to $ax^{4}+by^{4}=z^{2}$ were $\mu(a)=0=\mu(b)$. By Proposition $\S$
2.3 we now that $\gcd(a,b)=1=\gcd(c,d)$. ∎

######  Proposition 4.2.

Let $2n\in\mathbb{N}$ were $n>1$. If there is a divisor $d$ of $2n$ so that
$(dx^{2})^{2}+((\frac{2n}{d}y^{2}))^{2}=z^{2}$ for some $d|2n$, then $n$ is
congruent.

###### Proof.

Let $2n\in\mathbb{N}$ were $n>1$ and assume there is a divisor $d$ of $2n$ so
that $(dx^{2})^{2}+((\frac{2n}{d}y^{2}))^{2}=z^{2}$ for some $d|2n$ and
$x,y\in\mathbb{Z}^{+}$. Rearrange this to look like
$(x(dx))^{2}+(y(\frac{n}{d}y))^{2}=z^{2}$. By Theorem $\S$ 2.2 it follows that
the number $\frac{1}{2}(\frac{dx}{y})(\frac{\frac{2n}{d}y}{x})$ is congruent.
This is precisely $n$. ∎

######  Corollary 4.1.

Let $p$ be prime. Then $p$ is congruent if and only if one of the equations

$x^{4}+4p^{2}y^{4}=z^{2},~{}4x^{4}+p^{2}y^{4}=z^{2}$

has a solution for some $(x,y,z)\in\mathbb{N}^{3}$.

###### Proof.

Suppose that $p$ is prime. In view of proposition $\S$ 4.2 if there is a
divisor $d$ of $2p$ so that $(dx^{2})^{2}+((\frac{2p}{d}y^{2}))^{2}=z^{2}$ for
some $d|2p$, then $p$ is congruent. Since $p$ is prime the only divisors are
$d=1,2$ or $d=p$. Each case results in the three equations respectfully

$x^{4}+4p^{2}y^{4}=z^{2},~{}4x^{4}+p^{2}y^{4}=z^{2},~{}p^{2}x^{4}+4y^{4}=z^{2}$

Two of the equations are the same so this reduces to the equations in the
hypothesis. Conversely, suppose one of these equations are satisfied. For any
equation there is a $d\in\\{1,2,p\\}$ for which we may rewrite this as
$(x(dx))^{2}+(y(\frac{2p}{d}y))^{2}=z^{2}$. Note that $x|dx$ and
$y|(\frac{2p}{d}y)$ so by Theorem $\S$ 2.2 the number
$\frac{1}{2}\left(\frac{dx}{y}\right)\left(\frac{(\frac{2p}{d}y)}{x}\right)$
is congruent. This is precisely $p$. ∎

######  Theorem 4.1.

Let $a,b\in\mathbb{N}$ were $a,b>1$ and $\gcd(a,b)=1$. Then the equation
$ax^{4}+by^{4}=z^{2}$ has a solution $(x,y,z)\in\mathbb{N}^{3}$ if and only if
the equation $a+b=z^{2}$ has a solution $z\in\mathbb{Z}$.

###### Proof.

Let $a,b>1$ and assume the equation $ax^{4}+by^{4}=z^{2}$ has a solution
$(x,y,z)\in\mathbb{N}^{3}$. Then we have the following implications

$\displaystyle ax^{4}+by^{4}$ $\displaystyle=z^{2}$ $\displaystyle\Downarrow$
$\displaystyle\frac{a}{b}x^{4}+y^{4}$ $\displaystyle=\frac{1}{b}z^{2}$
$\displaystyle\Downarrow$ $\displaystyle\frac{a}{b}x^{4}$
$\displaystyle=\frac{1}{b}z^{2}-y^{4}$ $\displaystyle\Downarrow$
$\displaystyle\frac{a}{b}$
$\displaystyle=\frac{\frac{1}{b}z^{2}-y^{4}}{x^{4}}$ $\displaystyle\Downarrow$
$\displaystyle\frac{a}{b}$ $\displaystyle=\frac{z^{2}-by^{4}}{bx^{4}}$

Now $b|by^{4}$ but $b\not|z^{2}$ since $\gcd(b,z)=1=\gcd(x,z)=\gcd(x,y)$. This
means that $\gcd(z^{2}-by^{4},bx^{4})=1=\gcd(a,b)$. In other words, the
fractions we found to be equal are actually in lowest terms. This implies that
$b=bx^{4}$ so then $x=\pm 1$. By analogy we have the implication

$\displaystyle ax^{4}+by^{4}$ $\displaystyle=z^{2}$ $\displaystyle\Downarrow$
$\displaystyle x^{4}+\frac{b}{a}y^{4}$ $\displaystyle=\frac{1}{a}z^{2}$
$\displaystyle\Downarrow$ $\displaystyle\frac{b}{a}y^{4}$
$\displaystyle=\frac{1}{b}z^{2}-x^{4}$ $\displaystyle\Downarrow$
$\displaystyle\frac{b}{a}$
$\displaystyle=\frac{\frac{1}{b}z^{2}-x^{4}}{y^{4}}$ $\displaystyle\Downarrow$
$\displaystyle\frac{b}{a}$ $\displaystyle=\frac{z^{2}-bx^{4}}{by^{4}}$

Again we see $a|ay^{4}$ but $a\not|z^{2}$ since
$\gcd(a,z)=1=\gcd(y,z)=\gcd(x,y)$. This means that
$\gcd(z^{2}-bx^{4},ay^{4})=1=\gcd(a,b)$. In other words, the fractions we
found to be equal are actually in lowest terms. This implies that $b=bx^{4}$
so then $x=\pm 1$. Hence, we have found $x=1=y$. So then $ax^{4}+by^{4}=z^{2}$
implies that $a+b=z^{2}$.
Secondly suppose that $a+b=z^{2}$ for some $z^{2}$. Then note that this is a
solution to $ax^{4}+by^{4}=z^{2}$ were $x=\pm 1=y$. This completes the proof.
∎

######  Theorem 4.2.

If there exist $a,b,c\in\mathbb{Z}$ so that
$\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ were as $\gcd(a,b,c)=1$, then the the
equation $a+b=z^{2}$ has a solution $z\in\mathbb{Z}$.

###### Proof.

Assume there exist a $a,b,c\in\mathbb{N}$ so that
$\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ were as $\gcd(a,b,c)=1$. Multiple
through by $abc$ to obtain $bc+ac=ab$. So then $c(b+a)=ab$. Let $\gcd(a,b)=d$
and set $a=a^{\prime}d$ and $b=b^{\prime}d$. Thus we have
$\frac{1}{d}(c(b+a))=\frac{1}{d}ab$. This is equivalent to
$c(b^{\prime}+a^{\prime})=b^{\prime}a^{\prime}d$. Since $a^{\prime}$, and
$b^{\prime}$ are relatively prime so it must be that
$(b^{\prime}+a^{\prime})|d$. But $d|(b^{\prime}+a^{\prime})$ so this means
that $(b^{\prime}+a^{\prime})=d$. Hence,
$(a+b)=(a^{\prime}d+b^{\prime}d)=d(a^{\prime}+b^{\prime})=d^{2}$. ∎

######  Proposition 4.3.

Each proper congruent number gives rise to a solution of the equation
$ax^{4}+y^{4}=z^{2}$ for some $a\in\mathbb{N}$.

###### Proof.

By Proposition $\S$ 4.1 each proper congruent number corresponds to an integer
solution to the equation $ax^{4}+by^{4}=z^{2}$ for some relatively prime
numbers $a$ and $b$. By Theorem $\S$ 4.2, since $x$ and $y$ cannot be $1$
(this follows since $n$ is proper) it follows that $a=1$ or $b=1$, but they
cannot both be $1$. In any either case were $a=1$ or $b=1$ we may rearrange
the equation to look like $ax^{4}+y^{4}=z^{2}$ or $x^{4}+by^{4}=z^{2}$. Since
addition is commutative and the fact that $x$ and $y$ are variables these
equations are the same. Thus each proper congruent number corresponds to an
integral solution to the equation $ax^{4}+y^{4}=z^{2}$. ∎

##  5 Semi-Proper Congruent Numbers

######  Definition 5.1.

Let $n$ be a congruent number which corresponds to triangle $\triangle ABC$
with rational sides $(\frac{a}{b},\frac{c}{d},\frac{e}{f})$. If $b=1$ or $d=1$
but not both, then we say that $n$ is a Semi-proper congruent number.

######  Proposition 5.1.

If $n$ is a semi proper congruent number with $\triangle ABC$ constructed by
the triple $(a,\frac{c}{d},\frac{e}{f})$, then $f=d$ and $e^{2}=c^{2}+ad^{2}$.

###### Proof.

Suppose $n$ is a semi proper congruent number with $\triangle ABC$ constructed
by the triple $(a,\frac{c}{d},\frac{e}{f})$. Since $n$ is the area of the
$\triangle ABC$ we have the identity $n=\frac{1}{2}(a)(\frac{c}{d})$ with
$\gcd(c,d)=1$. Thus $d\not|c$ but this means that then $d|a$. Write $a=dk$ for
some $k\in\mathbb{N}$. Thus

$\displaystyle(\frac{c}{d})^{2}+a^{2}$ $\displaystyle=(\frac{e}{f})^{2}$
$\displaystyle\Downarrow$ $\displaystyle(\frac{c}{d})^{2}+(dk)^{2}$
$\displaystyle=(\frac{e}{f})^{2}$ $\displaystyle\Downarrow$ $\displaystyle
c^{2}+d^{4}k^{2}$ $\displaystyle=d^{2}(\frac{e}{f})^{2}$

Recall that $\gcd(e,f)=1$ by assumption. Since
$c^{2}+d^{4}k^{2}=d^{2}(\frac{e}{f})^{2}$ and the left side of the equation is
a positive whole number it must follows that $f|d$. Since $f|d$ there exist a
$l$ such that $d=fl$. Hence, we have the following implication:

$\displaystyle(\frac{c}{d})^{2}+a^{2}$ $\displaystyle=(\frac{e}{f})^{2}$
$\displaystyle\Downarrow$ $\displaystyle(\frac{c}{fl})^{2}+a^{2}$
$\displaystyle=(\frac{e}{f})^{2}$ $\displaystyle\Downarrow$
$\displaystyle(\frac{c^{2}}{(fl)^{2}}+\frac{a^{2}(fl)^{2}}{(fl)^{2}}$
$\displaystyle=(\frac{e}{f})^{2}$ $\displaystyle\Downarrow$
$\displaystyle(\frac{c^{2}+a^{2}(fl)^{2}}{(fl)^{2}})$
$\displaystyle=(\frac{e}{f})^{2}$

By assumption we know that $\gcd(e,f)=1$. Therefore, the right side of the
expression is in lowest terms. Since $d=fl\not|c$ it follows that the left
side of the expression is in lowest term. Two quotients in lowest terms have
equal numerator and denominator. Hence, $f^{2}=f^{2}l^{2}$ which implies
$l=1$. Since $f=dl$ and $l=1$ we find $f=d$ and $e^{2}=c^{2}+ad^{2}$ as
desired. ∎

######  Theorem 5.1.

If $a,c,e,d$ are integers that satisfy the following conditions:
(i) $e^{2}=c^{2}+(da)^{2}$
(ii) $\gcd(c,d)=1$ and $d|a$
(iii) $\\{a,c\\}$ contains an even integer, and this even number has more
factors of 2 than $2d$ has.
Then the number $\frac{1}{2}(a)(\frac{c}{d})$ is a Semi-proper congruent
number.

###### Proof.

Let conditions $(i)-(iii)$ hold valid. Set $f=d$, then we have
$(\frac{e}{f})^{2}=(\frac{c}{d})^{2}+a^{2}$. This is a Pythagorean triple
which represents a triangle with area $\frac{1}{2}(a)(\frac{c}{d})$. Thus this
integer is a Semi-proper congruent number. ∎

######  Proposition 5.2.

Let $a>1$ and suppose $x^{2}+a^{2}y^{4}=z^{2}$ for some $x,y\in\mathbb{N}$.
Then the number $n=\frac{1}{2}ax$ is a Semi-Proper congruent number.

###### Proof.

Suppose that $a>1$ and $x^{2}+a^{2}y^{4}=z^{2}$ for some $x,y\in\mathbb{N}$.
Rewrite this as $x^{2}+(y(ay))^{2}=z^{2}$. Since this is a Pythagorean triple
by proposition $\S$ 2.3 it follows that $\gcd(x,y(ay))=1$. Since $y|ay$ and at
least one of $\\{x,ay\\}$ is even conditions $(i)-(iii)$ of Theorem $\S$ 5.1
are satisfied. As a consequence the number
$\frac{1}{2}(ay)(\frac{x}{y})=\frac{1}{2}ax$ is semi-congruent. ∎

##  6 Improper Congruent Numbers

This section is short because there is not to much to prove or show. Improper
congruent numbers are simple congruent numbers because they represent the are
of a right triangle whose sides are specifically whole integers and not
rational numbers. I call these simple because Euclid has already described
these type of numbers for us.

######  Definition 6.1.

A Improper Congruent number is a number $n\in\mathbb{N}$ so that $n$ is the
area of some right triangle $\triangle ABC$ with integer sides $(a,b,c)$ That
is all sides lengths of the $\triangle ABC$ are strictly in $\mathbb{Z}^{+}$
and not in $\mathbb{Q}$.

######  Theorem 6.1.

Let $k\in\mathbb{N}$. If $k$ is a improper congruent number then it can be
written as $mn(m-n)(m+n)$ for some $m,n\in\mathbb{Z}^{+}$ such that
$\gcd(m,n)=1$, $m>n$ and $m\not\equiv n(\bmod{2})$. Moreover, $mn(m-n)(m+n)$
is also some improper congruent number.

###### Proof.

By Theorem $\S$ 2.1 it follows that all primitive Pythagorean triples can be
written as

$\displaystyle a$ $\displaystyle=m^{2}-n^{2}$ $\displaystyle b$
$\displaystyle=2nm$ $\displaystyle c$ $\displaystyle=m^{2}+n^{2}$

were $\gcd(n,m)=1$ and $m>n$ and $m\not\equiv n(\bmod{2})$. The area of each
such triangle is thus $k=\frac{1}{2}2mn(m-n)(m+n)=mn(m-n)(m+n)$ ∎

##  7 Characterizing Congruent Numbers Using the Mobius function

In this section I show how to generate all the types of congruent number using
Pythagorean triples. We have already done two examples of this in Ex. $\S$
2.1, and Ex. $\S$ 2.1. I will show that all congruent numbers can be written
as

$\frac{nm(m-n)(m+n)}{\left(\tau\sigma\right)^{2}}$

were as $\tau|\varrho(2nm)$ and $\sigma|\varrho((m-n)(m+n))$ for some $m,n$. I
will show that Improper congruent numbers correspond to this representation
when the values $d_{1}=1=d_{2}$, Semi-proper congruent numbers correspond to
only one value $\tau=1$ or else $\sigma=1$. Moreover, I demonstrate that each
proper congruent number corresponds to values $\tau\neq 1\neq\sigma$. In order
to accomplish this I use the preliminaries that were established throughout
this work. As a consequence we develop some useful results about congruent
numbers. Specifically, these results show that some of that the divisors $d$
of $\varrho(mn(m-n)(m+n)$ and every congruent number corresponds to one of
these divisors.

######  Theorem 7.1.

Every Improper-Congruent number can be written as $m^{3}n-n^{3}m$ whereas $m$
is not square free, $n$ is not square free,or possibly when both $m$ and $n$
are not square free, and were as $\gcd(m,n)=1$, $m>n$, and $m\neq
n(\bmod{2})$.

###### Proof.

Suppose that $\alpha$ is an Improper-Congruent number. By definition $\alpha$
is the area of a right triangle $\triangle ABC$ with all sides $\\{a,b,c\\}$
all in $\mathbb{Z}^{+}$. By Theorem $\S$ 2.1 there exist $m$ and $n$ such that

$\displaystyle a$ $\displaystyle=m^{2}-n^{2}$ $\displaystyle b$
$\displaystyle=2nm$ $\displaystyle c$ $\displaystyle=m^{2}+n^{2}$

The area of this triangle $\triangle ABC$ is
$\frac{1}{2}\left(m^{2}-n^{2}\right)\left(2nm\right)=m^{3}n-n^{3}m$. ∎

######  Theorem 7.2.

Every Semi-proper congruent number can be written as

$\frac{mn(m-n)(m+n)}{\sigma^{2}}$

were as $\sigma|\varrho(nm(m-n)(m+n))$,and $\gcd(m,n)=1$, $m>n$, and
$m\neq(\bmod{2})$ with $\sigma|nm$ or $\sigma|(m-n)(m+n)$.

###### Proof.

The proof follows by an argument of decent. Suppose that $\alpha$ is an Semi-
Proper congruent number corresponding to the triple
$\\{a,\frac{c}{d},\frac{e}{f}\\}$ with and $\mu(\alpha)=1$. Then by
proposition $\S$ 5.1 it follows that $e^{2}=c^{2}+a^{2}d^{2}$. Note that this
is a solution to the equation $x^{2}+a^{2}y^{2}=z^{2}$ with $x=a$, $y=d$, and
$z=e$. Now suppose that $\mu(y)=0$. Then there exist a $d|y$ such that
$d^{2}|y$. Hence, write $y=d^{2}k_{1}$ and notice that the equation
$x^{2}+a^{2}y^{2}=z^{2}$ transforms into the equation
$x^{2}+(ak_{1})^{2}d^{4}=z^{2}$. Note that this is a solution to the equation
$x^{2}+a^{2}y^{4}=z^{2}$. Thus by direct application of proposition $\S$ 5.2
it follows that the number $\frac{1}{2}xak_{1}$ is a Semi-Congruent number.
Now we apply the same argument again. If $\mu(d)=0$ then there exist a
$\tau|d$ such that $\tau^{2}|d$. Therefore we may write $d=\tau^{2}k_{2}$ for
some $k_{2}\in\mathbb{Z}$. Then by direct substitution the equation
$x^{2}+(ak_{1})^{2}d^{4}=z^{2}$ transforms into the equation

$x^{2}+(ak_{1}k_{2})^{2}\tau^{4}=z^{2}$

and again by proposition $\S$ 5.2 it follows that $\frac{1}{2}xak_{1}k_{2}$ is
a Semi-Congruent number. We take note that if $k_{1}\neq 1\neq k_{2}$ then
$\frac{1}{2}xak_{1}<\frac{1}{2}xak_{1}k_{2}$, which implies that these are two
distinct Semi-Congruent numbers. By repeating this argument until we reach a
point were the equation $x^{2}+a^{2}y^{4}=z^{2}$ has $\mu(y)=0$ we find a set
of Semi-Congruent numbers, which all correspond to this single Pythagorean
triple. In complete generality each congruent number we find looks like
$\frac{1}{2}x\tau$ for some $\tau|(ay)^{2}$ with $\mu(t)=0$. Hence, since
$\tau$ is a divisor of the number $(ay)^{2}$ it must divide $ay$. By Euclid’s
theorem $\S$ 2.1 we have $ay=(m-n)(m+n)$ or else $ay=2nm$ for some $m$ and $n$
relatively prime and of opposite parity. Hence, $\tau$ must divide
$(m-n)(m+n)$ or else $2nm$. But then $\tau$ can be written as
$\frac{2mn}{\sigma^{2}}$ or $\frac{(m-n)(m+n)}{\sigma^{2}}$ for some
$\sigma|(m-n)(m+n)$ or $\sigma|2mn$. In either case the are of such a triangle
is

$\frac{1}{2}x\tau=\frac{1}{2}\frac{2mn(m-n)(m+n))}{\sigma^{2}}=\frac{nm(m-n)(m+n)}{\sigma^{2}}$

In the second case when $mu(y)=1$ we find $\sigma=1$ yields the result. Since,
this holds for each divisor $\tau$ with $\mu(t)=0$ the result of the
hypothesis then follows. ∎

######  Theorem 7.3.

Every Proper congruent number can be written as

$\frac{mn(m-n)(m+n)}{(\sigma\tau)^{2}}$

were as $\sigma^{2}|\varrho(mn)$, $\tau|\varrho((m-n)(m+n))$ and
$\gcd(m,n)=1$, $m>n$, and $m\neq n(\bmod{2})$.

###### Proof.

The proof of this theorem is similar to the proof of the last two theorems.
The proof is summarized and not given rigorously to make sure the reader has
paid close attention thus far. First suppose that $\alpha$ is a Proper-
Congruent number. Then we want to show that it can be written in the form
presented in the hypothesis. First since $\alpha$ is Proper it represents the
area of some triangle corresponding to the triple
$\\{\frac{a}{b},\frac{c}{d},\frac{e}{f}\\}$. So then
$e^{2}=(da)^{2}+(bc)^{2}$. Now this a Pythagorean triple which corresponds a
triangle with areas $\frac{1}{2}\frac{a}{b}\frac{c}{d}$. The trick here is we
can reformulate $d$, $a$, $b$, and $c$ by simply writing the factorization of
them all. However, we notice that of we shift the prime factorization of $d$
into $a$, meaning we take some prime from $d$ and give them to $a$ and write
$da=(\tau a)$ were $\frac{a}{d}=\tau$. Then we see that $e^{2}=(\tau
a)^{2}+(bc)^{2}$. As a consequence of Theorem $\S$ 2.1 it follows that the
number $n=\frac{1}{2}\frac{a}{b}\frac{c}{\tau}$ is also a Proper-Congruent
number. Clearly, the quantities $\frac{1}{2}\frac{a}{b}\frac{c}{d}$ and
$\frac{1}{2}\frac{a}{b}\frac{c}{\tau}$ are not the same if $\tau\neq d$. Now
we can repeat the argument until we exhaust all the non square free divisors
of $a$, and then use the same argument for the quantity $b$, and both $a$ and
$b$ simultaneously. By doing this we find the result, by finding the area of
each each one of these triangles. These argument is summarized by noting that
the number of congruent number that correspond to each Pythagorean triple is
equal to the number of non square free divisors of for its components $a$ and
$b$ corresponding to the triple $\\{a,b,c\\}$. ∎

######  Theorem 7.4.

Let $A$ be given as

$A=\\{n\in\mathbb{N}:\text{n is a proper congruent number }\\}$

and $B$ be given as

$B=\\{(a,b,c)\in\mathbb{N}^{3}:a^{2}+b^{2}=c^{2},\mu(a)=0=\mu(b)\\}$

Let $(\alpha,\beta,\gamma)\in B$. and write $\alpha$, and $\beta$ in canonical
form:

$\alpha=\prod_{p|a}{p^{e_{i}}},\indent\beta=\prod_{p|b}{p^{e_{i}^{\prime}}}$

Then the number of congruent numbers that correspond to the triple
$(\alpha,\beta,\gamma)$ is given by the quantity

$\sum_{i=1}^{\max(k,k^{\prime})}{\lfloor\frac{e_{i}+e_{i}^{\prime}}{2}\rfloor}$

were $k$ and $k^{\prime}$ denotes the number of prime factors of $\alpha$ and
$\beta$ respectfully.

###### Proof.

This proof is left to the reader as an exercise. (Hint! Use Table 3.1 to help
understand the hypothesis, then explain why each non-square divisor of
$\alpha$ and $\beta$ corresponds to a unique congruent number and after add
them both.) ∎

## References

  * [1] PLozano-Robledo, Alvaro, and Alvaro Lozano-Robledo. Elliptic curves, modular forms, and their L-functions. Providence, RI: American Mathematical Society, 2011.
  * [2] Conrad, Keith. ”The congruent number problem.” The Harvard College Mathematics Review 2 (2008): 58-74.
  * [3] Fitzpatrick, Richard. Euclid’s Elements In Greek. Lulu.com, 2006
  * [4] Lozano-Robledo, Alvaro, and Alvaro Lozano-Robledo. Elliptic curves, modular forms, and their L-functions. Providence, RI: American Mathematical Society, 2011.
  * [5] Koblitz, Neal I. Introduction to elliptic curves and modular forms. Vol. 97. Springer Science & Business Media, 2012.
  * [6] Wiles, A. ”The Birch and Swinnerton-Dyer Conjecture, Clay Mathematics Institute Web Site.”
  * [7] J. Tunnell: A Classical Diophantine Problem and Modular Forms of Weight 3/2, Invent. Math.72(1983),323–334
  * [8] Conrad, Keith. ”The congruent number problem.” The Harvard College Mathematics Review 2 (2008): 58-74.
  * [9] Tian, Ye. ”Congruent numbers with many prime factors.” Proceedings of the National Academy of Sciences 109.52 (2012): 21256-21258.


# A Generalized Kalman Filter Augmented Deep-Learning based Approach for
Autonomous Landing in MAVs
††thanks: NG acknowledges an NVIDIA GPU grant and Associateship sponsored by
the International Center for Theoretical Physics
978-1-6654-3323-5/21/$31.00 ©2021 IEEE

Pranay Mathur1 Dept. of EEE
BITS Pilani, K.K Birla Goa Campus
Goa, India
<EMAIL_ADDRESS>Yash Jangir2 Dept. of EEE
BITS Pilani, K.K Birla Goa Campus
Goa, India
<EMAIL_ADDRESS>Neena Goveas3 Department of Computer
Science
BITS Pilani, K.K Birla Goa Campus
Goa, India
<EMAIL_ADDRESS>
###### Abstract

Autonomous landing systems for Micro Aerial Vehicle (MAV) have been proposed
using various combinations of GPS based, vision and fiducial tag-based
schemes. Landing is a critical activity that a MAV performs and poor
resolution of GPS, degraded camera images, fiducial tags not meeting required
specifications and environmental factors pose challenges. An ideal solution to
MAV landing should account for these challenges and for operational challenges
which could cause unplanned movements and landings. Most approaches do not
attempt to solve this general problem but look at restricted sub-problems with
at least one well-defined parameter. In this work, we propose a generalized
end-to-end landing site detection system using a two-stage training mechanism,
which makes no pre-assumption about the landing site. Experimental results
show that we achieve comparable accuracy and outperform existing methods for
the time required for landing.

###### Index Terms:

Autonomous aerial vehicles, Deep-learning, Kalman Filter, Robot Operating
System

##  I Introduction

Recent years have seen the rapid development and maturity of Micro Aerial
Vehicle (MAV) technology as they are agile, easily maneuverable and offer cost
effective solutions. The constraints of cost and weight in MAV has meant that
many of the techniques developed for larger Unmanned Aerial Vehicles (UAV)
cannot be used as is. One of the main concerns in MAV research is to design
safe landing techniques in unfavorable situations where GPS based precise
localization is not feasible. Vision based techniques which are commonly used,
have to work within the constraints of a non ideal camera and image capture
and account for partial or complete occlusion of the target site during the
landing operation.

In this work, we propose a novel two-stage trained deep learning-based method
that achieves accurate vision-based landing. There are no assumptions made
regarding the landing site or any symbols placed on it. A short training time
ensures our model can also be be used in situations where the image of the
landing site are available just prior to flight. We have successfully tested
our solution during landing operations where there was occlusion and image
distortion.

We find that our proposed technique gives comparable performance in terms of
accuracy and outperforms other solutions in terms of the time required for
landing. The contributions of this paper are as follows:

  * •

An implementation and training strategy for a deep-learning framework that
allows passing images of an arbitrary landing site prior to flight and makes
no assumptions on the contours, profile or texture of this site.

  * •

A customized Kalman Filter based controller that utilises the constraints
imposed by the bounding box coordinates to achieve precise control during
landing.

In addition, we have also created a custom dataset for training our network
initially prior to using augmented dataset of the actual landing site images
for the second stage of training.

##  II Related Works

UAVs have found their way into various applications involving military,
delivery, search and rescue operations. Six-degrees of freedom and their agile
behaviour make them an ideal tool for critical operations in inaccessible
regions. For many missions, certain scenarios lead to further constraints in
the form of lack of infrastructure, requiring smaller and cheaper devices[1].
MAVs are cost-effective and agile but lack of on-board computational resources
and peripherals, results in the need for customized software which can ensure
smooth functioning. One of the basic requirements for a MAV is autonomous
landing in a precise manner without depending on external support, for
example, a positioning method based on the global positioning system (GPS)[2]
or pre-decided image tags to be kept on the landing site. Autonomous landing
using images captured by on-board cameras is one of the options which has
provided promising results[3].

There exist a host of previous research work attempting to tackle the issue of
vision-guided autonomous MAV landing. A broad categorization can be performed
on the basis of whether the visual sensor used is on-board[4, 5, 6] or off-
board[7, 8, 9]. We focus on on-board visual system approaches, which are of
two kinds, classical image processing techniques such as contour recognition
and corner detection [10, 11, 12] or learning based techniques [13, 14].

Autonomous vision based landing for MAVs consists of two phases, first
involves identification of landing sites and the second involves visual
estimation containing position estimation and direction estimation to generate
a path towards the landing site. Issues affecting the detection of the landing
site include the fact that the captured image of the landing site is subject
to affine transformation, scale change, rotation, translation and oblique
projection transformation[15]. Moreover environmental factors such as
vibration and uneven illumination give rise to degradation of the quality of
images captured. Most techniques make strong assumptions regarding the
illumination of the environment and the contours and profile of the tags used.
An ideal approach should perform well in the above non-ideal circumstances and
cases where occlusion of the landing site occurs.

The issues related to direction estimation and correction process during the
landing operation is typically controlled by a Kalman fiter based approach.
Existing methods that use Kalman filters with inputs from classical vision
based techniques have multiple drawbacks, such as the tracking error
accumulating [16]. Most proposed techniques make assumptions that all details
of the landing site are known well in advance and the MAV will not encounter
any unknown parameter. This results in loss of flexibility and imposing
infrastructural requirements on the end-user.

In this work we focus on on-board vision based solutions for an arbitrary
landing site with no assumptions regarding image capturing system. Our
approach is able to handle scenarios where partial occlusion of the landing
site occurs and the landing vehicle field of view includes only partial views
of the landing site in the captured images.

##  III Methodology

In this study, an end-to-end deep-learning strategy is proposed to enable on-
board vision based landing for arbitrary landing sites. Our approach consists
of the following - the first phase where we deploy a target bounding box
detection network to extract coordinates of the landing site in an image
captured by the MAV. This is explained in Section III-A. The deep-learning
based method has a two-stage training strategy which is detailed in Section
III-B. Our method is designed for high-speed and accurate detection of
arbitrary landing sites. In the second phase, we use these detected
coordinates and a customized Kalman Filter based controller to perform
landing. This is detailed in Section III-C.

###  III-A Landing Site Detection

Identification of the landing site in images captured by the MAV is a critical
step for performing autonomous landing. We pre-process our images by
conversion to 8-bit grayscale images and use CLAHE[17], an equalization
technique that tackles the problem of over-amplification of the contrast and
makes our system more robust to changes in environmental illumination. Instead
of using a pre-defined fiducial tag or set of known landing sites we use an
arbitrary image of the landing site. This is enabled by our model architecture
detailed in Section III-A1 and our customized training strategy in Section
III-B.

####  III-A1 Model Architecture

To achieve the goal of fast detection of an arbitrary image of the landing
site available only prior to the flight, we are constrained in the total time
available for training of our network. We solve this by pre-training our
network and utilizing transfer learning.

We employ the Faster R-CNN[18] model architecture for accurate and fast
estimation of the bounding-box coordinates. It comprises of two modules, a
deep fully convolutional network that proposes regions, referred to as the
Region Proposal Networks(RPN), and a Fast R-CNN[19] detector that extracts
features from the proposed regions. This technique is appropriate for our case
in the context of Transfer Learning where we are constrained to train only on
a small dataset, in our particular case the actual landing site images, using
the weights of a network trained on a larger sized dataset, our custom
dataset, detailed in IV-A. The input images from the MAV are represented as
$Height\times Width\times Depth$ tensors, which are passed through a pipeline
consisting of a pre-trained Convolutional Neural Network layers and end with a
convolutional feature map. This is used as a feature extractor for the RPN.
The Region Proposal Network uses the features that the CNN computes to find
probable landing sites in the image. This number can be pre-defined with an
upper bound. The RPN uses fixed sized reference bounding boxes, referred to as
anchors, placed uniformly throughout the original images. These are the
regions checked for containing relevant objects and their location is noted.
The features extracted by the CNN are used and Region of Interest (RoI)
Pooling is applied to relevant locations. The final module is the R-CNN
module, which uses the previous information to classify the content in the
bounding box and adjust the bounding box coordinates. Finally a post-
processing step to prevent duplicate proposals is applied. This algorithmic
approach is referred to as Non-Maximum Suppression (NMS). These bounding-box
coordinates are then passed to our controller.

We find that our choice of the network architecture gives superior performance
in-terms of accuracy after transfer learning, size of required dataset,
required time for training, and time for inference.

###  III-B Training

Training is an essential step to tune the weights of a model framework for
accurate inference but poses a major challenge when the landing site images
are received just prior to flight. This lack of time imposes constraints on
number of epochs that can be used for training. Additionally, we are
constrained to perform this training on the on-board computer and not on a
computationally powerful server.

We propose a novel two stage strategy to solve this problem. In the first
stage we train our network on our custom dataset detailed in IV-A. This
dataset contains a wide variety of possible landing sites in a variety of non-
ideal scenarios and environmental conditions and is split in a 70:20:10 ratio
of training, validation and test data. Initially, we train the network on our
custom dataset till we achieve a $loss$ tending to $unity$. This is done to
adjust and provide an accurate initial estimate of the graph weights. This
coarse tuning is performed in order to achieve a reduction in the time
required for training on the actual site without compromising on accuracy. The
second stage involves finer adjustment of the weights by training our model
pre-flight and is performed using images of the actual site. Image
augmentation is used to expand the size of the dataset and improve
performance. This involves horizontal and vertical shift, random rotation,
random zoom and random brightness augmentation. Training is performed on this
dataset using the weights of the graph obtained from the previous step till
the $loss$ tends to 0.5. We find that this heuristically determined value is a
limit below which we get accurate results. This stage presents a trade-off
between training times and resultant accuracy. These have been discussed in V.

###  III-C Kalman Filter based Controller

We use a customized Kalman filter based controller built upon the work of [20]
for precisely maneuvering the quadcopter to the landing site. In our method,
instead of using fiducial tags for landing site detection[20], we use Kalman
filter based controller with bounding-box coordinates from our model
architecture. A bounding-box gives comparatively less information about pose
of the landing site than is received from fiducial tags. Additionally, in case
of partially visible or occluded landing sites, fiducial tag based methods
cannot work as the tag must be fully visible for detection. In contrast, our
controller will use the best available coordinates and thus will work even in
case of partially occluded images.
Our method consists of passing the bounding box coordinates of the landing
site detected and transforming bounding-box coordinates to states, to act as
an input to the kalman filter. Resulting predicted states are passed to state
controller which calculates errors and generates subsequent motion control
commands for the MAV as shown in Fig.1.

Figure 1: A block diagram of working

In Fig.3 :

$\overline{H}$ ${}:{}$ |  Detected Observations by Deep learning Model
---|---
$B_{t}$ ${}:{}$ |  Detected Corners
$Z_{t}$ ${}:{}$ |  Transformed states
$\widehat{X_{t}}$ ${}:{}$ |  Predicted states by Kalman Filter
$U_{t}$ ${}:{}$ |  Generated motion control signals by state controller

####  III-C1 Transformation of Bounding Box Coordinates to States

We receive the bounding box coordinates of the target landing site from the
first stage. We define these bounding box coordinates as $(X_{i},Y_{i})$ for
$i\in\\{0,1,2,3\\}$.

These coordinates are used to calculate the states, $Z_{t}$, which will be
passed to the Kalman Filter. These states consist of the centroid coordinates
of bounding box, $X_{c}$ and $Y_{c}$, the angle between the landing site and
horizontal of image, $\theta$ and the width and height of the landing site
$W_{s}$ & $H_{s}$ respectively. For the Angle between the landing site and
horizontal of image ($\theta$), we calculate Vectors :

$\begin{split}\overrightarrow{X}=\overrightarrow{X_{1}}-\overrightarrow{X_{0}}\\\
\overrightarrow{Y}=\overrightarrow{Y_{1}}-\overrightarrow{Y_{0}}\end{split}$
(1)

where ($X_{0},Y_{0}$) and ($X_{1},Y_{1}$) are bounding box coordinates
$(X_{i},Y_{i})$ for $i\in\\{0,1\\}$

Calculation of the angle between these two vectors gives us the value of
$\theta$.
If $E_{P_{1},P_{2}}$ is euclidean distance between any two points in 2-D
Cartesian coordinate system then $W$ is calculated by averaging two parallel
euclidean widths of box and $H$ calculated by averaging two parallel euclidean
heights of box.

$\begin{split}W=\dfrac{E_{B_{0},B_{1}}+E_{B_{2},B_{3}}}{2}\\\
H=\dfrac{E_{B_{0},B_{2}}+E_{B_{1},B_{3}}}{2}\end{split}$ (2)

####  III-C2 State Controller

We use a Proportional Integral Derivative Controller(PID)[21] to navigate to
the target states with feedback from the predicted states by the Kalman
Filter. Predicted States corresponding target landing site in the image are
received periodically from Kalman Filter. Unlike aruco tags or any other
fiducial tags where a complete view of the tag would be required, in our
method even in cases where the landing site is partially visible, the Kalman
Filter uses the available coordinates to accomplish a successful landing.

To control of X,Y and $\theta$, We define centroid coordinates of image frame
as :

$X_{i_{c}}$ ${}:{}$ |  X Coordinate of the centroid of image frame
---|---
$Y_{i_{c}}$ ${}:{}$ |  Y Coordinate of the centroid of image frame

Closed loop PID controllers are used to control the position along X and Y
axis and the orientation with respect to landing site. In order to achieve the
target landing site position,the image frame centroid, $X_{i_{c}},Y_{i_{c}}$
must coincide with the landing site centroid prediction by Kalman Filter. PID
controllers generate appropriate velocity commands for the Autopilot using the
landing site centroid coordinates for feedback.

In aruco tags and other fiducial tag based methods extra information in the
form of pose of the tag which includes orientation is received, however this
information is absent in bounding box coordinates.
In our method for orientation, Angle between the landing site and horizontal
of image, $\theta$ which is calculated by bounding box coordinates, is used.
Since we use a bounding box which has a symmetric orientation on rotation of
$90^{o}$, the closest direction parallel to the edges is considered 0.
$\theta$ is directly the error for PID controller. In order to reach the
correct orientation $\theta$ must tend to zero. Hence, PID controller
generates control signals as yaw rate commands for the MAV to achieve correct
orientation with respect to the target landing site.

Figure 2:  Closed Loop PID controller

Control of altitude $Z$: While the position along X and Y axis and the
orientation are controlled by PID controllers, the altitude cannot directly be
controlled by a PID controller. As the MAV descends down, the width, $W$ and
height, $H$ of the target landing site increases as going down will increase
the size of object as seen in camera image. This shows a dependency of
altitude on $W$ and $H$. So, we estimate this dependency in altitude as:

$\begin{split}Z_{e}=K*|W-H|\end{split}$ (3)

where $K$ represents the proportionality constant.

We define a condition based on error in altitude,$Z_{e}$ and current altitude
of MAV, $Z_{i}$. While this condition is satisfied, the MAV continues
descending by a predefined altitude, $Z_{f}$. We generate new target altitude
position, $Z_{pos}$ for the MAV:

$\begin{split}Z_{pos}=Z_{i}-Z_{f}\end{split}$ (4)

Hence, we are able to generate motion control commands with the predicted
states of the target landing site, the MAV aligns itself with landing site and
descends down using the State controller to reach to the landing site.

####  III-C3 Final Landing Sequence

In order to land safely on the target site, the MAV must be in the correct
position, orientation and altitude, with respect to the target landing site.
At the instant the pose aligns with the target landing site and the altitude
is below a pre-defined threshold of 0.3 m, landing sequence is initiated by
the State controller. The state controller terminates publishing control
commands and descent at a controlled rate in a linear trajectory is carried
out.

##  IV Implementation

We implement and evaluate our proposed solution in the Gazebo Simulation
environment using the Robot Operating System (ROS) framework[22]. The
infrastructure and hardware simulated is ubiquitously available and our
implementation is agnostic to hardware specification.

###  IV-A Dataset

We use a custom built dataset to train our model on. These are obtained
primarily from simulations carried out in gazebo. The images are collected
across a variety of environmental conditions such as excessive lighting,
differential lighting, partial occluded sites and across different landing
sites including simple boxes and white pads against backgrounds such as grass
patches, roads and dirt patches. The dataset contains 450 images. A
representative subset of the images collected to counter the problem of
merging backgrounds with minute color and contrast difference have been shown
in Fig. 4. We highlight the fact that we do not intend to train the model
solely for this particular dataset, merely alter and give an initial estimate
of the weights in the graph so training may be faster on images of the actual
landing site. An accurate initial estimate of weights are achieved using
comparatively fewer number of images when compared to a conventional dataset.

###  IV-B Simulation Environment

We use the Gazebo simulator to perform evaluations of our approach. Gazebo
enables us to perform realistic simulations using superior rendering and an
accurate physics engine. Varied parameters such as illumination in the
environment, landing sites, surface against which the landing site is
simulated and the number of objects present in the vision cone are simulated.
An example of a simulated environment is shown in Fig. 3 We use the PX4
framework[23] which is an open source flight control software. It is highly
portable, Operating System independent and supports a variety of peripherals.
Mavros, a MAVLink extendable communication node for ROS is used for
communication with PX4. A DJI F450 frame, equipped with a GPS, a pixhawk 2.4.8
flight controller and a downward facing camera publishing images at 20 $Hz$
with a resolution of $640\times 480$ is used.

Figure 3: Simulated environment in Gazebo

Figure 4: Representative images from the Dataset

##  V Evaluation Studies

The algorithm is evaluated on the x86 architecture using a dual-core Intel
Core i3-4010U CPU with a processor clocked at 1.7 GHz along with a 8GB DDR4
RAM on a laptop. These specifications are comparable to typical computers used
on-board a MAV. In the experiments, we evaluate the performance of our
approach on five different categories of landing sites and scenarios,$L_{n}$,
(Table I). The illumination scale varies from 1-10 according to environmental
lighting used in Gazebo with 1 being extremely low lighting to 10 being
extremely bright.

TABLE I: Landmarks Site | Illumination | Background | Length(m) | Breadth(m)
---|---|---|---|---
$L_{1}$ | 2 | Road | 1.8 | 0.5
$L_{2}$ | 3 | Road | 0.7 | 0.8
$L_{3}$ | 4 | Plain | 1.2 | 1
$L_{4}$ | 7 | Grass | 0.5 | 0.5
$L_{5}$ | 9 | Road | 1.1 | 0.8

Few images of the actual landing site are augmented to produce a dataset for
the second stage of training, of size 25 images in our experiment. Using the
weights available after our first stage of training, the model is further
trained with the above images. The number of epochs for which the model is
trained is shown in Table II. On our processor, these times varied from an
average of $623s$ for 50 epochs to an average of $1837s$ for 100 epochs. Our
experiments investigate the robustness of our approach in different
environmental illumination intensities and against different backgrounds while
recording accuracy and computational usage parameters. We highlight an
assumption made here that the accuracy is enough to ensure the target landing
site comes into the field-of-view of the downward facing camera. Our results
show that we achieve landing the MAV on certain significantly smaller
landmarks in adverse environmental conditions. A summary of the results of has
been shown in Table II.

The average frame processing time increased in ideal illumination and was
significantly lower in dimly lit environments. A similar trend was observed in
CPU usage, while in terms of accuracy, the highest was obtained in scenario
$L_{4}$ shown in Fig. 6 where illumination was adequate. The lowest accuracy
was in $L_{1}$ proving among visual factors, illumination proves to be the
most critical. Apart from this even in cases of partial occlusion and in cases
where an affine transformation existed due to the orientation of the MAV, the
landing site was detected properly. In addition to scenario $L_{4}$ which had
multiple visually different objects visible, we also test against multiple
similar objects visible in the frame. These cases have been shown in Fig.7. As
shown in the $top-left$ and $top-right$ image, the detector was trained for 50
epochs on two different landing sites and detects each of them accurately. A
standard sequence of steps involved in landing has been shown in Fig.5.

TABLE II: Accuracy, Timing ,CPU Usage and Training Time Site | Accuracy (%) | Avg. Time (ms) | CPU (%) | Epochs
---|---|---|---|---
$L_{1}$ | 92.1 | 62.11 | 72.78 | 50
$L_{2}$ | 97.5 | 62.64 | 75.23 | 50
$L_{3}$ | 96.8 | 66.56 | 63.70 | 70
$L_{4}$ | 98.4 | 74.12 | 74.13 | 80
$L_{5}$ | 95.2 | 70.62 | 77.89 | 100

A comparison of our method with a GPS based method and fiducial tag based
methods [3] and [24] has been shown in table III. While the average accuracy
is comparable, we out-perform the method in-terms of average landing time and
landing speed.

TABLE III: Comparative Table Source | Accuracy (m) | Landing Speed (m/s) | Time(s)
---|---|---|---
Ours | 0.16 | 0.25 | 41.7
GPS-based | 1-3 | 0.6 | -
Wubben et al. [3] | 0.11 | 0.1 | 55
Araar et al. [24] | 0.13 | 0.06 | -

Figure 5: Sequence of frames showing Quadcopter landing

Figure 6: Site detection - Obstacles and Grass Background

Figure 7: Corner Cases in Landing Site detection

##  VI Conclusion

In this work we have introduced a novel deep learning vision based MAV
autonomous landing system augmented by a Kalman Filter. We implement a
training strategy that allows passing the landing site prior to flight and our
approach achieves high accuracy comparable to standard deep-learning based
methods where the landing site is fixed. We implement a Kalman Filter that
utilises the constraints imposed by the bounding box coordinates to achieve
precise control for landing on the site. Our method works even on partially
occluded sites and in a diverse set of environmental conditions. Future work
includes predicting landing paths with geometrically tight trajectory
constraints, implementation on moving platforms and adding the additional
application of obstacle avoidance using the downward facing camera.

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