
3. Find all pairs of natural numbers $x, y$ such that $\frac{x y^{2}}{x+y}$ is a prime.


Solution. Assume that the natural numbers $x, y$ and a prime $p$ satisfy the equation

$$
\frac{x y^{2}}{x+y}=p
$$

Denote by $d$ the greatest common divisor of $x$ and $y$. Then $x=d a$ and $y=d b$, where $a$ and $b$ are relatively prime. Substituting into (1), clearing the denominators and dividing by the positive number $d$, we obtain

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

Since the numbers $a$ and $b$ are relatively prime, so are the two numbers $b^{2}$ and $a+b$ from the two sides of the equality (2). It follows ${ }^{1}$ that $b^{2} \mid p$. The prime $p$ has only two divisors: 1 and $p$, of which the second is not a square; thus necessarily $b=1$. Substituting this into (2) gives

$$
d^{2} a=p(a+1)
$$

Arguing again as above, i.e. since $a$ is a divisor of the left-hand side of (3), it is also a divisor of the right-hand side, and as $a, a+1$ are relatively prime, we conclude that $a \mid p$. Thus either $a=1$, or $a=p$. We discuss these two cases separately.

If $a=1,(3)$ reads $d^{2}=2 p$. As $p$ is a prime, $2 p$ can be a square only if $p=2$. Then also $d=2$ and we arrive at the first solution $x=d a=2, y=d b=2$.

If $a=p$, then dividing by the positive number $p$ (3) becomes $d^{2}=p+1$, or $p=(d+1)(d-1)$. As $p$ is a prime and $d-1<d+1$, necessarily $d-1=1$ and $d+1=p$. This gives $d=2, p=3$, and we have arrived at the second solution $x=d a=d p=6$ and $y=d b=2$.

It is easy, though not necessary (all the manipulations we performed were equivalent), to check that both pairs are indeed solutions of the original equation (1).

Answer: there are exactly two such pairs $(x, y)$, namely, $(2,2)$ and $(6,2)$.
