1. Determine the integer solutions of the equation $p(x+y)=xy$, where $p$ is a given prime number.
Solution. Note that the pair $(0,0)$ is a solution to the problem and that the pair $(x, y)$ in which exactly one of the numbers is equal to 0 is not a solution to the problem. Let $p(x+y)=x y \neq 0$. Then $p \mid x y$, and since $p$ is a prime number, it follows that $p \mid x$ or $p \mid y$. Let, for example, $x=m p$, where $m$ is an integer and $m \neq 0$. It is easy to see that in this case $m$ cannot be $1$. Further, $p(m p+y)=m p y$, so $y=\frac{m p}{m-1}$.

1) If $m-1=1$, then $x=2 p, y=2 p$ is a solution to the given equation.
2) If $m-1=-1$, then $m=0$, which contradicts the assumption.
3) If $m-1=p$, then $x=p(p+1), y=p+1$ is a solution to the given equation.
4) If $m-1=-p$, then $x=p(1-p), y=p-1$ is a solution to the given equation.
5) If $m-1 \notin\{-1,1,-p, p\}$, then $y=\frac{m p}{m-1}$ is not an integer.

Similarly, the case when $y$ is divisible by $p$ is considered, and in this case, we obtain two more solutions: $x=p+1, y=p(p+1)$ and $x=p-1, y=p(1-p)$.