Being p a prime number prov that the equation 1/x + 1/y = 1/p has exactly 3 solutions, for x,y naturals

Clearing fractions yields p(x + y) = xy.
==> xy - px - py = 0
==> xy - px - py + p^2 = p^2
==> (x - p)(y - p) = p^2.

Since x,y,p are positive integers, and p is prime, we have only 3 possibilities:
(i) x - p = 1 and y - p = p^2 ==> (x, y) = (p + 1, p^2 + p)
(ii) x - p = p and y - p = p ==> (x, y) = (2p, 2p)
(iii) x - p = p^2 and y - p = 1 ==> (x, y) = (p^2 + p, p + 1).

I hope this helps!