#  The Rational Number $\mathbf{\frac{n}{p}}$ as a sum of two unit fractions

Konstantine Zelator
Department of Mathematics, Statistics, and Computer Science
212 Ben Franklin Hall
Bloomsburg University of Pennsylvania
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Bloomsburg, PA 17815
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P.O. Box 4280
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##  1 Introduction

In a 2011 paper in the journal Asian Journal of Algebra (see [1]), the authors
consider, among other equations, the diophantine equations

$2xy=n(x+y)\ \ {\rm and}\ \ 3xy=n(x+y).$

For the first equation, with $n$ being an odd positive integer, they give the
solution (in positive integers $x$ and $y$)
$\dfrac{n+1}{2}=x=\dfrac{(n-1)}{2}+1$,
$y=n\left(\dfrac{(n-1)}{2}+1\right)=n\left(\dfrac{n+1}{2}\right)$.

For the second equation, with $n\equiv 2({\rm mod}3)$, they present the
particular solution,

$\dfrac{n+1}{3}=x=\dfrac{(n-2)}{3}+1,\ \
y=n\left(\dfrac{(n-2)}{3}+1\right)=n\left(\dfrac{n+1}{3}\right).$

If in the above equations we assume $n$ to be prime, then these two equations
become special cases of the diophantine equation, $nxy=p(x+y)$, with $p$ being
a prime and $n$ a positive integer with $n\geq 2$.

This two-variable symmetric diophantine equation is the subject matter of this
article; with the added condition that the integer $n$ is not divisible by the
prime $p$. Observe that this equation can be written equivalently in fraction
form:

$\dfrac{n}{p}=\dfrac{1}{x}+\dfrac{1}{y}.$

This problem then can be approached from the point of view of decomposing a
positive rational number into a sum of two unit fractions (i.e., two rational
numbers whose numerators are equal to $1$). The ancient Egyptians left behind
an entire body of work involving the decomposition of a given fraction into a
sum of two or more unit fractions. They did so by creating tables containing
the decomposition of specific fractions into sums of unit fractions. An
excellent source on the subject of the work of the ancient Egyptians on unit
fractions is the book by David M. Burton, “The History of Mathematics, An
Introduction” (see [2]). Note that thanks to the identity
$\dfrac{1}{k}=\dfrac{1}{k+1}+\dfrac{1}{k(k+1)}$, a unit fraction can always be
written as a sum of two unit fractions.

We state our theorem.

######  Theorem 1.

Let $p$ be a prime, $n$ a positive integer, $n\geq 2$. Also, assume that
gcd$(p,n)=1$ (equivalently, $n$ is not divisible by $p$). Consider the two-
variable symmetric diophantine equation,

$nxy=p(x+y)$ (1)

with the two variables $x$ and $y$ taking values from the set
${\mathbb{Z}}^{+}$ of positive integers. Then,

  1. (i)

If $n=2$ and $p\geq 3$, equation (1) has exactly three distinct solutions, the
following positive integer pairs:

$(x,y)=(p,p),\ (x,y)=\left(p\left(\dfrac{p+1}{2}\right),\
\dfrac{p+1}{2}\right),$

and its symmetric counterpart

$(x,y)=\left(\dfrac{p+1}{2},\ p\left(\dfrac{p+1}{2}\right)\right).$

  2. (ii)

If $n\geq 3$, and $n$ is a divisor of $p+1$. Then equation (1) has exactly two
distinct solutions:

$(x,y)=\left(p\left(\dfrac{p+1}{n}\right),\ \dfrac{p+1}{n}\right)\ {\rm and}\
(x,y)=\left(\dfrac{p+1}{n},\ p\left(\dfrac{p+1}{n}\right)\right).$

  3. (iii)

If $n$ is not a divisor of $p+1$, Equation (1) has no solution.

##  2 A lemma from number theory

The following lemma, commonly referred to as Euclid’s lemma, is of great
significance in number theory.

######  Lemma 1.

(Euclid’s lemma): Suppose that $a,b,c$ are positive integers such that $a$ is
a divisor of the product $bc$; and gcd$(a,b)=1$ (i.e., $a$ and $b$ are
relatively prime), then $a$ must be a divisor of $c$.

Typically, this lemma and its proof can be found in an introductory number
theory book. For example, see reference [3].

##  3  Proof of Theorem 1

First we show that the positive integer pairs listed in Theorem 1 are indeed
solutions to Equation (1).

If $n=2$ and $p\geq 3$, then for $(x,y)=(p,p)$, a straightforward calculation
shows both sides of (1) are equal to $2p^{2}$; and for
$(x,y)=\left(\dfrac{p(p+1)}{2},\dfrac{p+1}{2}\right)$, a calculation shows
that both sides of (1) are equal to $\dfrac{p(p+1)^{2}}{2}$.

If $n\geq 3$ and $n$ is a divisor of $p+1$, then for
$(x,y)=\left(p\left(\dfrac{p+1}{n}\right),\dfrac{p+1}{n}\right)$, a
calculation shows that both sides of equation (1) are equal to
$\dfrac{p(p+1)^{2}}{n}$.

In the second part of this proof, we show that there are no other solutions to
equation (1). To do so, we will demonstrate that if $(t_{1},t_{2})$ is a
solution to (1), then it must be one of the solutions listed in Theorem 1\.
So, let $(t_{1},t_{2})$ be a positive integer solution to equation (1).

We have,

$\left\\{\begin{array}[]{c}p(t_{1}+t_{2})=nt_{1}t_{2}\\\ \\\
t_{1},t_{2}\in{\mathbb{Z}}^{+}\end{array}\right\\}$ (2)

Let $d$ be the greatest common divisor of $t_{1}$ and $t_{2}$. Then

$\left\\{\begin{array}[]{l}t_{1}=du_{1},\ t_{2}=du_{2};\\\ {\rm for\
relatively\ prime\ positive\ integers}\ u_{1}\ {\rm and}\ u_{2};\\\ {\rm
gcd}(u_{1},u_{2})=1\end{array}\right\\}$ (3)

From (2) and (3) we obtain,

$p(u_{1}+u_{2})=nd\,u_{1}u_{2}$ (4)

Since the prime $p$ is relatively prime to $n$. By (4) and Lemma 1, it follows
that $p$ must divide the product $du_{1}u_{2}$. Since $p$ is a prime number,
it must divide at least one of $d$ and $u_{1}u_{2}$. We distinguish between
two cases: The case wherein $p$ divides the product $u_{1}u_{2}$; and the case
in which $p$ is a divisor of $d$.

Case 1: $p$ is a divisor of $u_{1}u_{2}$.

Since $p$ is a prime, and the integers $u_{1}$ and $u_{2}$ are relatively
prime by (3), and also in view of the fact that $p$ divides the product
$u_{1}u_{2}$, it follows that $p$ must divide exactly one of $u_{1},u_{2}$. It
must divide one but not the other. Thus, there are two subcases in Case 1.

Subcase 1a being the one with $p|u_{1}$ (i.e., $p$ divides $u_{1}$);

Subcase 1b: $p$ divides $u_{2}$.

But these two subcases are symmetric since equation (4) is symmetric in
$u_{1}$ and $u_{2}$. Thus, without loss of generality, we need only consider
the subcase $p|u_{1}$. So we set

$\left(u_{1}=pv_{1},\ v_{1}\ {\rm a\ positive\ integer}\right)$ (5)

Combining (5) with (4) we get,

$\left\\{\begin{array}[]{c}pv_{1}+u_{2}=nd\,v_{1}u_{2}\\\ \\\ {\rm or\
equivalently},\ u_{2}=v_{1}\cdot(ndu_{2}-p)\end{array}\right\\}$ (6)

According to (6), the positive integer $v_{1}$ is a divisor of $u_{2}$. But,
by (5) $v_{1}$ is also a divisor of $u_{1}$. Since $u_{1}$ and $u_{2}$ are
relatively prime by (3), it follows that

$v_{1}=1$ (7)

Hence, by (7) and (6), we further obtain,

$p=u_{2}(nd-1)$ (8)

According to (8), $u_{2}$ is a divisor of $p$, and since $p$ is a prime it
follows that either $u_{2}=1$ or $u_{2}=p$. If $u_{2}=1$, then (8) yields
$p+1=nd$ which implies that $n$ is a divisor of $p+1$. Using
$d=\dfrac{p+1}{n},\ v_{1}=1,u_{2}=1$, we also get $u_{1}=p$ (by (5)). So, by
(3) we obtain the solution
$t_{1}=p\left(\dfrac{p+1}{n}\right),t_{2}=\dfrac{p+1}{n}$ (already a verified
solution in the first part of the proof). Now, if $u_{2}=p$ in (8), then
$2=nd$ which implies either $n=2$ and $d=1$, or $n=1$ and $d=2$. But $n\geq
2$, so the latter possibility is ruled out. Thus, $u_{2}=p,\ n=2$, and $d=1$.
Also, by (7) we have $v_{1}=1$ and so $u_{1}=p$ by (5).

Hence, (3) yields $t_{1}=p=t_{2}$; $(p,p)$ with $n=2$ being a solution
verified in the first part of the proof.

Case 2: $p$ is a divisor of $d$

We set

$(d=p\delta,\ \delta\ {\rm is\ a\ positive\ integer})$ (9)

by (9) and (4) we have,

$u_{1}+u_{2}=n\delta u_{1}u_{2}$ (10)

Clearly, by inspection, we see that equation (10) implies that the positive
integers $u_{1}$ and $u_{2}$ must divide each other. Since they are relatively
prime, it follows that

$u_{1}=u_{2}=1$ (11)

Equations (10) and (11) yield

$2=n\delta$ (12)

Due to the fact $n\geq 2$, (12) implies that $n=2$ and $\delta=1$. So, by
(11), (9), and (3), it is clear that (since $d=p$) $u_{1}=u_{2}=p$. This
produces $(u_{1},u_{2})=(p,p)$, with $n=2$. An already verified solution. The
proof is complete. $\Box$

## References

  * [1] Kishan, Hari, Rani, Megha and Agarwal, Smiti, The Diophantine Equations of Second and Higher Degree of the Form $3xy=n(x+y)$ and $3xyz=n(xy+yz+zx)$, etc., Asian Journal of Algebra 4(1), (2011), pp. 31-37.
  * [2] Burton, David M., “The History of Mathematics, An Introduction”, Sixth Edition, McGraw Hill, (2007), p. 40.
  * [3] Rose, Kenneth H., “Elementary Number Theory and Its Applications”, 5th Edition, Pearson, Addison Wesley, (2005), p. 109.