## Task A-3.2.

Determine all pairs of natural numbers $x$ and $y$ for which $\frac{x y^{2}}{x+y}$ is a prime number.
## Solution.

If $d=M(x, y)$, we can write $x=d a, y=d b$, where $a, b \in \mathbb{N}, M(a, b)=1$.

Let $p=\frac{x y^{2}}{x+y}$. Then $p=\frac{d a \cdot(d b)^{2}}{d a+d b}$.

It follows that

$$
d^{2} a b^{2}=p(a+b)
$$

Since $M(a, a+b)=M(b, a+b)=M(a, b)=1$, we conclude that $a b^{2}$ divides $p$, which is only possible if $b=1$ (the equation becomes $d^{2} a=p(a+1)$) and $a=1$ or $a=p$.

We distinguish two cases:

1. $a=1$

Then $d^{2}=2 p$. Since $p$ is a prime number, it follows that $p$ divides $d$, and thus $p^{2}$ divides $d^{2}=2 p$. From this, we conclude that $p^{2}$ divides $2 p$, so $p$ divides 2. Therefore, $p=2$ and $d=2$. The only solution in this case is $(x, y)=(2,2)$.
2. $a=p$

Then $d^{2}=p+1$, or $(d-1)(d+1)=p$. Since $p$ is a prime number and $0 \leqslant d-1< d+1$, it follows that $d-1=1$ and $d+1=p$. From this, we get $d=2$ and $p=3$, so the only solution in this case is $(x, y)=(6,2)$.

Finally, the desired pairs are $(x, y) \in\{(2,2),(6,2)\}$.