All That Reminder Is Gold

It is not too hard to rewrite the expression on the LHS as

$10-n+\dfrac{n+17}{n^2-2n-11}$

For this to be a prime number it must certainly be an integer!

Now a necessary condition (not a sufficient one) for this to be an an integer is that the numerator of the fraction is greater than the denominator. Think of the numerator as a line and the denominator as a concave upwards parabola. They intersect when

$n+17=n^2-2n-11 \\ \implies n^2-3n-28=0 \\ \implies(n+4)(n-7)=0$

The parabola lies below the line between $n=-4$ and $n=7$ and so we need to check the formula only for the 12 values of n satisfying $-4\le n \le 7$

This is easy to do using the TABLE function on a scientific calculator, and the only prime is $\boxed{7}$ occurring when $n=-2$

Alex Burgess points out that we should also test $n=-17$ . Then we get $10--17=27=3\times9$ so we are stll ok (whew!)

In order to narrow down the applicable $n$ , let us first solve for the polynomial quotient and remainder:

$p = \dfrac{n^3 -12n^2 +8n+93}{11+2n-n^2} = (-n+10) + \dfrac{n+17}{n^2 - 2n -11}$

Since $p$ and $n$ are both integers, the remainder $r$ , $\dfrac{n+17}{n^2 - 2n -11}$ must be integer as well.

The most obvious one is $n=-17$ , yielding a remainder of $0$ .

Then for such remainder to be an integer, the numerator $|n+17|$ must be greater than or equal to the denominator $|n^2 - 2n -11|$ :

$|n+17| \geq |n^2 - 2n -11|$

For $n < -17$ , $|n+17| = -n-17$ . On the other hand, $|n^2 - 2n -11| = |(n-1)^2 -12|$ ; the absolute value is positive for $n < -\sqrt{12}$ or $n > \sqrt{12}$ .

Thus, for $n < -17$ , the equation obtained would be: $-n-17 \geq n^2-2n-11$ . Then $0 \geq n^2-n+6 = (n-0.5)^2 + 5.75$ , with no real answer. Thus, there is no answer for $n < -17$ .

Hence, the possible values lie in $n > -17$ , for, first, $|n^2 - 2n -11| = n^2 - 2n -11$ for $n < -\sqrt{12}$ or $n > \sqrt{12}$ or, second, $|n^2 - 2n -11| = -n^2 + 2n +11$ for $-\sqrt{12}<n < \sqrt{12}$ .

For the latter scenario, we can similarly solve the equation $n+17 \geq -n^2+2n+11$ : $n^2-n+6 >0$ , which is true for all $n$ . Yet, the applicable range is $-\sqrt{12}<n < \sqrt{12}$ , or more specifically $-3 \leq n \leq 3$ .

Then for the former scenario, $n + 17 \geq n^2 - 2n -11$ . Then $0 \geq n^2 -3n -28 = (n-7)(n+4)$ . The applicable range is $-4 \leq n \leq 7$ and $n < -\sqrt{12}$ or $n > \sqrt{12}$ . Hence, the overall applicable range is $-4 \leq n \leq -3$ or $3 \leq n \leq 7$ .

Combining all possible ranges, we will obtain $-4 \leq n \leq 7$ or $n = -17$ . By trial and error, we can calculate the remainders $r$ :

For $n= -17$ , $r = 0$ .

For $n= -4$ , $r = 1$ .

For $n= -3$ , $r = \dfrac{7}{2}$ .

For $n= -2$ , $r = -5$ .

For $n= -1$ , $r = -2$ .

For $n= 0$ , $r = \dfrac{-17}{11}$ .

For $n= 1$ , $r = \dfrac{-3}{2}$ .

For $n= 2$ , $r = \dfrac{-19}{11}$ .

For $n= 3$ , $r = \dfrac{-5}{2}$ .

For $n= 4$ , $r = -7$ .

For $n= 5$ , $r = \dfrac{11}{2}$ .

For $n= 6$ , $r = \dfrac{23}{13}$ .

For $n= 7$ , $r = 1$ .

We can include the integer remainders to check which would satisfy a prime $p$ :

For $n= -17$ , $r = 0$ ; $p = 27$ .

For $n= -4$ , $r = 1$ ; $p = 15$ .

For $n= -2$ , $r = -5$ ; $p = 7$ .

For $n= -1$ , $r = -2$ ; $p = 9$ .

For $n= 4$ , $r = -7$ ; $p = -1$ .

For $n= 7$ , $r = 1$ ; $p = 4$ .

Finally, $p = 7$ is the only solution. Thus, such sum is $\boxed{7}$ .