, 2014

We substitute $k = n + 2$ such that
$$\begin{aligned} S(n) &= (k-2)^2 + (k-1)^2 + k^2 + (k+1)^2 + (k+2)^2 = 5k^2 + 10 = 5(k^2 + 2). \\ P(n) &= (k-2)^2(k-1)^2 k^2 (k+1)^2 (k+2)^2 = k^2(k^2-1)^2(k^2-4)^2. \end{aligned}$$
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 $k^2 \equiv -2 \pmod{k^2+2}$,
$$P(n) \equiv -2(-2-1)^2(-2-4)^2 \equiv -2^3 \cdot 3^4 \pmod{k^2+2}.$$
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 $2 \cdot 3^4$:
$$\begin{array}{c|cccccccccc} k^2+2 & 1 & 2 & 3 & 6 & 9 & 18 & 27 & 54 & 81 & 162 \\ k^2 & -1 & 0 & 1 & 4 & 7 & 16 & 25 & 52 & 79 & 160 \\ k & 0 & \pm 1 & \pm 2 & \pm 4 & \pm 5 & & & & & \end{array}$$
We conclude that $n = k - 2 \in \{-7, -6, -4, -3, -2, -1, 0, 2, 3\}$ are the only solutions. $\square$