The Rational Number n/p as a sum of two unit fractions

TLDR: In this paper, the authors considered the diophantine equations 2xy=n(x+y) and 3xy =n(n+1)/2 with the added condition that the intager n is not divisible by the prime p. Theorem 1.

Abstract: In a 2011 paper published in the journal "Asian Journal of Algebra"(see reference[1]), the authors consider, among other equations,the diophantine equations 2xy=n(x+y) and 3xy=n(x+y). For the first equation, with n being an odd positive integer, they give the solution x=(n+1)/2, y=n(n+1)/2. For the second equation they present the particular solution, x=(n+1)/3,y=n(n+1)/3, where is n is a positive integer congruent to 2modulo3. 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) (1), with p being a prime and n a positive integer greater than or equal to 2. This 2-variable symmetric diophantine equation is the subject matter of this article; with the added condition that the intager n is not divisible by the prime p. Observe that this equation can be written in fraction form: n/p= 1/x + 1/y(See [2] for more details) In this work we prove the following result, Theorem1(stated on page2 of this paper):Let p be a prime, n a positive integer at least2, and not divisible by p. Then, 1)If n=2 and p is an odd prime, equation (1) has exactly three distinct positive integer solutions:x=p, y=p ; x=p(p+1)/2, y=(p+1)/2 ; x=(p+1)/2, y=p(p+1)/2 2)If n is greater than or equal to 3, and n is a divisor of p+1. Then equation (1) has exactly two distinct solutions: x=p(p+1)/n, y=(p+1)/n ; x=(p+1)/n, y=p(p+1)/n 3) if n is not a divisor of p+1. Then equation (1) has no positive integer solution. The proof of this result is elementary, and only uses Euclid's Lemma from number theory,and basic divisor arguments(such that if a prime divides a product of two integers; it must divide at least one of them).