Hellenic Mathematical Olympiad ARCHIMEDES

Let $d=\gcd (x,y)$. Then there exist $a,b\in \mathbb{Z}$ such that $x=da$, $y=db$, $(a,b)=1$. By substitution to the given equation we get: $$\frac{da(db{)}^3}{da+db}=\frac{d^{3}a{b}^3}{a+b}=p.(1)$$ From $(a,b)=1$, we get $(a,a+b)=1$ and $(b^3,a+b)=1$, giving from relation (1) that $a+b\mid d^3$. We write $$\frac{d^3}{a+b}=k,(2)$$ where $k$ is a positive integer. Then (1) becomes: $ka{b}^3=p$, and hence $b^3\mid p$. Therefore $b=1$ and $ka=p$. Hence we have the following cases: (i) If $k=p,a=1$, then (2) becomes $\frac{d^3}{2}=p\Rightarrow 2p=d^3\Rightarrow 2\mid d$. Hence $8\mid d^3$ and $8\mid 2p$, absurd. (ii) If $k=1,a=p$. Then (2) becomes: $$d^3=p+1\Rightarrow d^3-1=p\Rightarrow (d-1)(d^2+d+1)=p.$$ Therefore, we get $d-1=1$, $d^2+d+1=p$ (since $d^2+d+1>d-1$), $$\iff d=2,p=7,\text{ and so: }(x,y,p)=(14,2,7).$$