121. Find all integer solutions of the equation $p(x+y)=xy$, where $p$ is a given prime number.
Solution. The given equation is equivalent to the equation

$$
x y-p x-p y+p^{2}=p^{2}
$$

i.e., to the equation $y(x-p)-p(x-p)=p^{2}$, or $(x-p)(y-p)=p^{2}$. The product of two integers $x-p$ and $y-p$ equals the square of a prime number $p$ only in the following cases:

1) $x-p=p, \quad y-p=p$;
2) $x-p=-p \quad y-p=-p$
3) $x-p=1, \quad y-p=p^{2}$;
4) $x-p=-1, \quad y-p=-p^{2}$;
5) $x-p=p^{2}, \quad y-p=1$;
6) $x-p=-p^{2} \quad y-p=-1$.

Accordingly, in these cases, we obtain six solutions to the given equation.

Answer: $x_{1}=2 p, y_{1}=2 p ; x_{2}=0, y_{2}=0 ; x_{3}=p+1, y_{3}=p+p^{2} ;$ $x_{4}=p-1, y_{4}=p-p^{2} ; x_{5}=p+p^{2}, y_{5}=p+1: x_{6}=p-p^{2}, y_{6}=p-1$.