## Problem 2.

Find all possible integer solutions, $x$ and $y$, of the equation:

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

where $p$ is a certain prime number.
## Solution:

From $p(x+y)=xy$ and the fact that $p$ is a prime number, it follows that $p$ divides $x$ or $y$. Since, in the statement, the roles of $x$ and $y$ are completely symmetric, we can, without loss of generality, assume that $p$ divides $x$ and that, consequently, there is a number $k$ such that

$$
x=kp
$$

Then, the proposed equation becomes

$$
p(kp+y)=kpy
$$

or,

$$
kp+y=ky
$$

Now, a few manipulations:

$$
kp+y=ky \Rightarrow 0=ky-kp-y=k(y-p)-y \Rightarrow p=k(y-p)-y+p=(k-1)(y-p)
$$

This shows that $k-1$ is a divisor of $p$ and, since $p$ is prime, there are four possibilities: i) That $k-1=-p$. We get $k=1-p$ and, therefore,

$$
x=(1-p)p, \quad y=p-1
$$

ii) That $k-1=-1$. It results that $k=0$, that is,

$$
x=0, \quad y=0
$$

iii) That $k-1=1$. Then $k=2$ and it results:

$$
x=2p, \quad y=2p
$$

iiii) That $k-1=p$. Then $k=p+1$ and it results:

$$
x=p(p+1), \quad y=p+1
$$

which are all the possible solutions of the proposed equation.