The sum of primes

The given expression can be factored as $(N - 17)(N^{2} - 197)$ . There are then $4$ possible options where this can yield a prime number:

• (i) $N - 17 = 1 \Longrightarrow N = 18$ , yielding $18^{2} - 197 = 127$ , which is indeed prime;

• (ii) $N - 17 = -1 \Longrightarrow N = 16$ , yielding $(-1)(16^{2} - 197) = -59$ , which is not prime, (as defined);

• (iii) $N^{2} - 197 = 1 \Longrightarrow N^{2} = 198$ , for which $N$ is not an integer;

• (iv) $N^{2} - 197 = -1 \Longrightarrow N^{2} = 196 \Longrightarrow N = \pm 14$ :

----- (a) $N = 14$ yields $(14 - 17)(14^{2} - 197) = (-3)(-1) = 3$ , which is prime;

----- (b) $N = -14$ yields $(-14 - 17)((-14)^{2} - 197) = (-31)(-1) = 31$ , which is also prime.

Thus there are three prime numbers that can be expressed in the given form, their sum being

$127 + 3 + 31 = \boxed{161}.$