Algebra or Number Theory?

Using long division for polynomials, we find that

$n^3 + 4n^2 + 4n - 14 = (n^2 + 2n + 2)(n + 2) + (-2n - 18)$

In order for $n^2 + 2n + 2$ to divide $n^3 + 4n^2 + 4n - 14$ , it must also divide the remainder:

$n^2 + 2n + 2 \mid (-2n - 18).$

The only way that this is possible is either when $| - 2n - 18| ≥ |n^2 + 2n + 2|$ or when $-2n-18 = 0$ . In the first case, this inequality only holds when $-4 \le n \le 4$ .

We test all $n$ within this range, and determine that the values of n which work are $n = -4, -2, -1, 0, 1, 4$ . In the second case, we additionally find that $n = -9$ works. Therefore, the sum is $\boxed{-11}$ .

Since $n^3 + 4n^2 + 4n - 14 \; = \; (n+2)(n^2 + 2n + 2) - 2(n+9)$ we see that $n^2 + 2n + 2$ divides $n^3 + 4n^2 + 4n - 14$ precisely when $n^2 + 2n + 2$ divides $2(n+9)$ , and so there must exist an integer $k$ such that $\begin{aligned} k(n^2+ 2n + 2) & = 2(n+9) \\ kn^2 + 2(k-1)n + 2(k-9) & = 0 \\ (kn + k-1)^2 & = (k-1)^2 - 2k(k-9) \; = \; 65 - (k-8)^2 \end{aligned}$ The only ways to write $65$ as a sum of two squares are $65 = 8^2 + 1^2 = 7^2 + 4^2$ . There are four cases to consider:

• If $|k-8| =1$ , then either $k=9$ and $|9n+8| = 8$ , or else $k=7$ and $|7n+6|=8$ . The only solution in the first case is $n=0$ , and the only solution in the second case is $n=-2$ .
• If |(|k-8|=4), then either $k=12$ and $|12n+11| = 7$ , or else $k=4$ and $|4k+3| = 7$ . The first case has no solutions, and the only solution of the second case is $n=1$ .
• If $|k-8|=7$ , then either $k=15$ and $|15n+14|=4$ , or else $k=1$ and $|n|=4$ . The first case has no solutions, and the second case has the solutions $n = \pm4$ .
• If $|k-8|=8$ , then either $k=16$ and $|16n+15|=1$ , or else $k=0$ , in which case the original equation is just $2(n+9)=0$ , and so $n=-9$ . The first case has $n=-1$ as a solution.

Thus the answer is $(-9) + (-4) + (-2) + (-1) + 0 + 1 + 4 = \boxed{-11}$ .