Austria 2014

We substitute $k=n+2$ such that $$\begin{array}{rl}S(n)& =(k-2{)}^2+(k-1{)}^2+k^2+(k+1{)}^2+(k+2{)}^2=5k^2+10=5(k^2+2).\\ P(n)& =(k-2{)}^2(k-1{)}^2{k}^2(k+1{)}^2(k+2{)}^2=k^2(k^2-1{)}^2(k^2-4{)}^2.\end{array}$$ As $P(n)$ is the square of the product of 5 consecutive integers, it is divisible by $5$. On the other hand, $k^2+2$ is not divisible by $5$, because $-2$ is a quadratic non-residue modulo $5$. Therefore, $S(n)$ divides $P(n)$ if and only if $k^2+2$ divides $P(n)$. As $k^2\equiv -2(\mathrm{mod}{k}^2+2)$, $$P(n)\equiv -2(-2-1{)}^2(-2-4{)}^2\equiv -2^3\cdot 3^4(\mathrm{mod}{k}^2+2).$$ Therefore, the assertion is equivalent to $k^2+2\mid 2^3\cdot 3^4$. However, $k^2+2$ is congruent to $2$ or $3$ modulo $4$. In particular, $4$ does not divide $k^2+2$. We conclude that the assertion is equivalent to $k^2+2\mid 2\cdot 3^4$. We now consider all positive divisors of $2\cdot 3^4$: $$\begin{array}{ccccccccccc}{k}^2+2& 1& 2& 3& 6& 9& 18& 27& 54& 81& 162\\ k^2& -1& 0& 1& 4& 7& 16& 25& 52& 79& 160\\ k& 0& \pm 1& \pm 2& \pm 4& \pm 5& & & & & \end{array}$$ We conclude that $n=k-2\in \{-7,-6,-4,-3,-2,-1,0,2,3\}$ are the only solutions. $\square$