Let p be a prime number. Prove that the equation $2/p=1/x+1/y$ has one solution with integers $0<x<y$.

Question

I wrote the equation down as $2z/pz=(y+x)/(xy)$, where z is an integer. This would create a fraction not in the most simplified version. $xy$ must have a greatest common factor $>1$ with z, since p is a prime. However, I am not sure how to proceed from here.

Answer 1

As in a comment above, we must have $p \neq 2.$

$2x>p, 2y>p.$ Multiply through by $2pxy$ to get $4xy-2px-2py =0,$ and $$ (2x-p)(2y-p) = p^2 $$ with $y > x,$ we get $1 \cdot p^2 = p^2$ and $x = \frac{p+1}{2}$ and $y = \frac{p^2 + p}{2}$

Answer 2

Every odd number (not just the odd primes) allows a solution :

$$\frac{2}{2n+1}=\frac{1}{n+1}+\frac{1}{2n^2+3n+1}$$