Problem 1574

Solution:
If $x=0$ then $y=p$ and if $y=0$ then $x=p$. We will show that there are no other solutions.

Suppose $x, y>0$. Since $x \neq y$, we have
$$p\left(x^{2}+x y+y^{2}\right)=(x+y)\left(x^{2}+y^{2}\right)$$
If $p$ divides $x+y$, then $x^{2}+y^{2}$ must divide $x^{2}+x y+y^{2}$ and so it must also divide $x y$. This is a contradiction as $x^{2}+y^{2} \geqslant 2 x y > x y$.

Thus $p$ divides $x^{2}+y^{2}$, so $x+y$ divides $x^{2}+x y+y^{2}$. As $x+y$ divides $x^{2}+x y$ and $y^{2}+x y$, it also divides $x^{2}$, $x y$ and $y^{2}$. Suppose $x^{2}=a(x+y)$, $y^{2}=b(x+y)$ and $x y=c(x+y)$. Then $x^{2}+x y+y^{2}=(a+b+c)(x+y)$, $x^{2}+y^{2}=(a+b)(x+y)$, while $(x+y)^{2}=x^{2}+y^{2}+2 x y=(a+b+2 c)(x+y)$ yields $x+y=a+b+2 c$.

Substituting into $(*)$ gives
$$p(a+b+c)=(a+b+2 c)(a+b)$$
Now let $a+b=d m$ and $c=d c_{1}$, where $\operatorname{gcd}\left(m, c_{1}\right)=1$. Then
$$p\left(m+c_{1}\right)=\left(m+2 c_{1}\right) d m$$
If $m+c_{1}$ and $m$ had a common divisor, it would divide $c_{1}$, a contradiction. So $\operatorname{gcd}(m, m+c_{1})=1$. and similarly, $\operatorname{gcd}\left(m+c_{1}, m+2 c_{1}\right)=1$. Thus $m+2 c_{1}$ and $m$ divide $p$, so $m+2 c_{1}=p$ and $m=1$. Then $m+c_{1}=d$ so $c \geqslant d=a+b$. Now
$$x y=c(x+y) \geqslant(a+b)(x+y)=x^{2}+y^{2}$$
again a contradiction.

Alternative Solution by PSC:
Let $d=\operatorname{gcd}(x, y)$. Then $x=d a$ and $y=d b$ for some $a, b$ such that $\operatorname{gcd}(a, b)=1$. Then
$$d^{4}\left(a^{4}-b^{4}\right)=p d^{3}\left(a^{3}-b^{3}\right)$$
which gives
$$d(a+b)\left(a^{2}+b^{2}\right)=p\left(a^{2}+a b+b^{2}\right)$$
If a prime $q$ divides both $a+b$ and $a^{2}+a b+b^{2}$, then it also divides $(a+b)^{2}-\left(a^{2}+a b+b^{2}\right)=a b$. So $q$ divides $a$ or $q$ divides $b$. Since $q$ also divides $a+b$, it must divide both $a$ and $b$. This is impossible as $\operatorname{gcd}(a, b)=1$. So $\operatorname{gcd}\left(a+b, a^{2}+a b+b^{2}\right)=1$ and similarly $\operatorname{gcd}\left(a^{2}+b^{2}, a^{2}+a b+b^{2}\right)=1$. Then $(a+b)\left(a^{2}+b^{2}\right)$ divides $p$ and since $a+b \leqslant a^{2}+b^{2}$, then $a+b=1$.

If $a=0, b=1$ then $(*)$ gives $d=p$ and so $x=0, y=p$ which is obviously a solution. If $a=1, b=0$ we similarly get the solution $x=p, y=0$. These are the only solutions.