Determine all prime numbers for which there exist strictly positive integers $x, y$ such that

$$
x\left(y^{2}-p\right)+y\left(x^{2}-p\right)=5 p
$$

## Exercise Solutions

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We start by factoring the left-hand term: we have

$$
x\left(y^{2}-p\right)+y\left(x^{2}-p\right)=x y(x+y)-p(x+y)=(x y-p)(x+y)=5 p
$$

Thus, since $x+y \geqslant 2$, we have 3 cases to consider:

- $x+y=5$ and $x y-p=p$. In this case, $x y=2 p$. Therefore, $(x, y)=(1,4),(2,3),(3,2)$, or $(4,1)$. We then obtain that $x y=4$ or $x y=6$. Thus, $p=2$ or $3$. If $p=2$, we verify that $x=1, y=4$ is indeed a solution because $(16-2)-4=10=2 p$. If $p=3$, we verify that $x=2, y=3$ is indeed a solution because $2 \times(9-3)+3 \times(4-3)=15=5 p$.

- $x+y=p$ and $x y-p=5$. In this case, $x y-(x+y)=5$, so $x y-x-y+1=6$. Therefore, $(x-1)(y-1)=6$, so, up to permutation, $(x-1, y-1)=(1,6)$ or $(2,3)$, so, up to permutation, $(x, y)=(2,7)$ or $(3,4)$. Thus, $p=2+7=9$ or $3+4=7$. Since $p$ is prime, $p=7$. Conversely, if $p=7$, we verify that $x=3, y=4$ is indeed a solution because $3(16-7)+4(9-7)=35=5 p$.

- $x+y=5 p$ and $x y-p=1$. By symmetry, we can assume $x \geqslant y$, so $x \geqslant 2.5 p$, thus $1=x y-p \geqslant 2.5 p-p=1.5 p$, which is absurd.

Therefore, the solutions are $p=2,3,7$.

## 2 Midway Training

## Statements