Find all prime numbers $p$ and all strictly positive integers $x, y$ such that $\frac{1}{x}-\frac{1}{y}=\frac{1}{p}$.
On:

$$
p(y-x)=x y
$$

We denote $: x=d a$ and $y=d b$ with $d=\gcd(x, y)$ and $\gcd(a, b)=1$.

It follows that:

$$
p(b-a)=d a b
$$

Thus, $a \mid p(b-a)$, and since $a$ and $b-a$ are coprime because $\gcd(a, b)=1$, by Gauss's lemma, we get $a \mid p$, i.e., $a=1$ or $a=p$. Similarly, $b=1$ or $b=p$. We note that $a>0$ and thus $a=1$ and $b=p$.

We obtain $d=p-1$ and $(p-1, p(p-1))$ is a solution for any prime $p$.