Problem 4. Find all pairs of natural numbers $x$ and $y$, such that the ratio $\frac{x y^{3}}{x+y}$ is a prime number.
4. Find all pairs of natural numbers $x$ and $y$ such that the ratio $\frac{x y^{3}}{x+y}$ is a prime number.

Answer: $x=14, y=2$.

Let $x y^{3}=p(x+y)$, where $p$ is a prime number. This means that one of the numbers $x$ and $y$ is divisible by $p$. Let's consider both cases.

First, assume that $y=m p$. Then $x m^{3} p^{2}=x+m p$. Since $p \geqslant 2$, we can write the chain of inequalities $x m^{3} p^{2} \geqslant 2 x m^{3} p \geqslant x p+m p > x+m p$.

Now, consider the case $x=k p$. After transformations, we get the equation $k\left(y^{3}-p\right)=y$. If $y^{3}-p \vdots d$ for some natural number $d$, then $y \vdots d$ and, consequently, $p \vdots d$, which means $d=1$ or $d=p$. We can take $d$ to be the number $y^{3}-p$ itself. Thus, either $y^{3}-p=p$, which is clearly impossible since in $y^{3}$ all prime factors appear at least to the third power; or $y^{3}-p=1$. In the latter case, we get $p=(y-1)\left(y^{2}+y+1\right)$, and due to the simplicity of $p$, it is necessary that $y=2$. Then $p=7$ and $x=14$.

$\pm$ The solution approach is generally correct, but the answer is lost. 5 points

$\pm$ The case $x \vdots p$ is considered. 4 points

- The case $y \vdots p$ is considered. 2 points