On the Diophantine Equation 1/a + 1/b = (q+1) / pq

TL;DR

This paper completely solves the Diophantine equation 1/a + 1/b = (q+1)/pq for integers a, b, where p and q are primes with q+1 dividing p-1, using elementary techniques.

Contribution

It provides a complete classification of solutions to a specific prime-related Diophantine equation with elementary methods.

Findings

All solutions are explicitly characterized.

The solutions depend on the divisibility condition q+1 | p-1.

The approach avoids advanced number theory techniques.

Abstract

Let $p$ and $q$ be distinct primes such that $q+1 | p-1$. In this paper we find all integer solutions $a$, $b$ to the equation $1/a + 1/b = (q+1)/pq$ using only elementary methods.

Equations

$$\frac{1}{a}+\frac{1}{b}=\frac{q+1}{pq},$$

$$a=\frac{\zeta p^2+pq}{q+1}\text{,}b=\frac{\zeta pq+q^2}{\zeta (q+1)}$$

$$a=\frac{\zeta pq+q^2}{\zeta (q+1)}\text{,}b=\frac{\zeta p^2+pq}{q+1},$$

Full text

1. Introduction

Problem A1 of the 2018 William Lowell Putnam Mathematical Competition asked competitors to find all ordered pairs of positive integers $(a,b)$ for which $1/a+1/b=3/2018$ [1]. This is an example of a Diophantine equation of the form $1/a+1/b=q+1/pq$, where $p$ and $q$ are distinct primes such that $q+1|p-1$. In this note we generalize this problem and obtain all integer solutions $a,b$ to the equation $1/a+1/b=q+1/pq$ for distinct primes $p,q$ such that $q+1|p-1$ using only elementary arithmetic.

Theorem 1**.**

Let

[TABLE]

where $p,q$ are distinct primes such that $q+1|p-1$. There are two trivial solutions to equation 1; namely $a=p,b=pq$, and $a=pq,b=p$. The remaining integer solutions of equation 1 are given by the following formulæ:

[TABLE]

and

[TABLE]

where $\zeta\equiv 1\mod q+1$ and $\zeta|q^{2}$.

Proof.

Rewrite equation 1 as $pq(a+b)=(q+1)ab$. Then $p|a$, $p|b$, or both. Consider first the case where $p|a$ and $p|b$. Write $a=pk$ and $b=pl$ for $k,l\in\mathbb{Z}^{+}$. Then $pq(pk+pl)=(q+1)p^{2}kl\implies q(k+l)=(q+1)kl$. If $k,l\geq 2$, this equation cannot be satisfied, so one of $k,l=1$. Suppose $k=1$, so $a=p$, and $q(1+l)=(q+1)l\implies l=q$, leading to the trivial solutions $a=p,b=pq$, and, mutatis mutandis, $a=pq,b=p$.

Suppose now that $p|a$ but $p\nmid b$. Again writing $a=pk$ for $k\in\mathbb{Z}^{+}$, we have $pq(pk+b)=(q+1)pkb\implies q(pk+b)=(q+1)kb$, or equivalently, $qpk=(qk+k-q)b$. Since $p\nmid b$, it must be the case that $p|qk+k-q$. Write $qk+k-q=\zeta p$ for $\zeta\in\mathbb{Z}^{+}$. Then $qk+k=\zeta p+q\implies(q+1)k=\zeta(p-1)+\zeta+q$ Since $q+1|p-1$, $\zeta\equiv 1\mod q+1$.

Solving for $k$ and then $a$ and $b$ above leads to the formulæ  given in equation 2. To obtain solutions in $\mathbb{Z}$, it must be the case that $q+1|\zeta p^{2}+pq$ and both $\zeta$ and $q+1$ divide $\zeta pq+q^{2}$. The first condition is always satisfied given our earlier assumption that $q+1|p-1$: writing $\zeta=\alpha(q+1)+1$, we have $\zeta p^{2}+pq=\alpha(q+1)p^{2}+p^{2}+pq=\alpha(q+1)p^{2}+p(p+q)$, and $p+q=p-1+q+1$ is divisible by $q+1$. It can similarly be shown that $q+1|\zeta pq+q^{2}$. However, $\zeta|\zeta pq+q^{2}$ if and only if $\zeta|q^{2}$, leading to the result in Equation 2. If instead we assume that $p|b$ and $p\nmid a$, then, mutatis mutandis, we arrive at the solutions given in Equation 3. ∎