Problem (Number Theory):
Determine all prime numbers $p$ and all positive integers $x$ and $y$ satisfying $x^{3}+y^{3}=$ $p(x y+p)$.

Solution:
Up to a swap of the first two entries, the only solutions are $(x, y, p)=(1,8,19)$, $(x, y, p)=(2,7,13)$ and $(x, y, p)=(4,5,7)$. The verification is routine.



Set $s=x+y$. Rewrite the equation in the form $s\left(s^{2}-3 x y\right)=p(p+x y)$, and express $x y$ :



$$

x y=\frac{s^{3}-p^{2}}{3 s+p} \tag{*}

$$



In particular,



$$

s^{2} \geq 4 x y=\frac{4\left(s^{3}-p^{2}\right)}{3 s+p}

$$



or



$$

(s-2 p)\left(s^{2}+s p+2 p^{2}\right) \leq p^{2}-p^{3}<0

$$



so $s<2 p$.



If $p \mid s$, then $s=p$ and $x y=p(p-1) / 4$ which is impossible for $x+y=p$ (the equation $t^{2}-p t+p(p-1) / 4=0$ has no integer solutions).



If $p \nmid s$, rewrite $(*)$ in the form



$$

27 x y=\left(9 s^{2}-3 s p+p^{2}\right)-\frac{p^{2}(p+27)}{3 s+p} .

$$



Since $p \nmid s$, this could be integer only if $3 s+p \mid$ $p+27$, and hence $3 s+p \mid 27-s$.



If $s \neq 9$, then $|3 s-27| \geq 3 s+p$, so $27-3 s \geq$ $3 s+p$, or $27-p \geq 6 s$, whence $s \leq 4$. These cases are ruled out by hand.



If $s=x+y=9$, then $(*)$ yields $x y=27-p$. Up to a swap of $x$ and $y$, all such triples $(x, y, p)$ are $(1,8,19),(2,7,13)$, and $(4,5,7)$.

Set again $s=x+y$. It is readily checked that $s \leq 8$ provides no solutions, so assume $s \geq 9$. Notice that $x^{3}+y^{3}=s\left(x^{2}-x y+y^{2}\right) \geq$ $\frac{1}{4} s^{3}$ and $x y \leq \frac{1}{4} s^{2}$. The condition in the statement then implies $s^{2}(s-p) \leq 4 p^{2}$, so $s<p+4$.



Notice that $p$ divides one of $s$ and $x^{2}-x y+y^{2}$. The case $p \mid s$ is easily ruled out by the condition $s<p+4$ : The latter forces $s=p$, so $x^{2}-x y+y^{2}=x y+p$, i. e., $(x-y)^{2}=p$, which is impossible.



Hence $p \mid x^{2}-x y+y^{2}$, so $x^{2}-x y+y^{2}=k p$ and $x y+p=k s$ for some positive integer $k$, implying



$$

s^{2}+3 p=k(3 s+p) \tag{**}

$$



Recall that $p \nmid s$ to infer that $3 k \equiv s(\bmod p)$. We now present two approaches.



1st Approach. Write $3 k=s+m p$ for some integer $m$ and plug $k=\frac{1}{3}(s+m p)$ into $(* *)$ to get $s=(9-m p) /(3 m+1)$. The condition $s \geq 9$ then forces $m=0$, so $s=9$, in which case, up to a swap of the first two entries, the solutions turn out to be $(x, y, p)=(1,8,19),(x, y, p)=(2,7,13)$ and $(x, y, p)=(4,5,7)$.



2nd Approach. Notice that $k=\frac{s^{2}+3 p}{3 s+p}=3+$ $\frac{s(s-9)}{3 s+p} \leq 3+\frac{1}{3}(s-9)=\frac{1}{3} s \leq \frac{1}{3}(p+3)$, since $s<p+4$. Hence $3 k \leq p+3$, and the congruence $3 k \equiv s$ $(\bmod p)$ then forces either $3 k=s-p$ or $3 k=s$.



The case $3 k=s-p$ is easily ruled out: Otherwise, $(* *)$ boils down to $2 s+p+9=0$, which is clearly impossible.



Finally, if $3 k=s$, then $(* *)$ reduces to $s=9$. In this case, up to a swap of the first two entries, the only solutions are $(x, y, p)=(1,8,19)$, $(x, y, p)=(2,7,13)$ and $(x, y, p)=(4,5,7)$.

Final answer: $(1,8,19), (2,7,13), (4,5,7)$