All Primes

Actually very nice problem. I will approach this problem using Modular Arithmetic.

Observe that $f(x)$ and $g(x)$ are odd $\forall x \in N$ . So, $p$ is an odd prime.

Let $x=p-k;k \in N$ . So, $x$ , $p$ and $k$ , $p$ are pairwise coprime integers.

So,

$f(x) \equiv -k^{5} +5k^{4} - 5k^{3} + 5k^{2} +1 \equiv 0 \mod p$ $g(x) \equiv -k^{5} +5k^{4} - 3k^{3} - 5k^{2} - 1 \equiv 0 \mod p$

Adding the both,

$-2k^{5} +10k^{4} - 8k^{3} \equiv -2k^{3} (k^{2} - 5k +4) \equiv 0\mod p$

Since, $p$ is an odd prime and is coprime to $k$ . So,

$k^{2} - 5k + 4 \equiv (k-1)(k-4) \equiv 0 \mod p$

So, either $k \equiv 4 \mod p$ or $k \equiv 1 \mod p$ .

Case 1 : If $k \equiv 4 \mod p$ , then $f(x) \equiv 17 \mod p$ and $g(x) \equiv -17 \mod p$ . So, $\boxed{p=17}$ only satisfies this for $x = 13$ .

Case 2 : If $k \equiv 1 \mod p$ , then $f(x) \equiv 5 \mod p$ and $g(x) \equiv -5 \mod p$ . So, $\boxed{p=5}$ only satisfies this for $x = 4$ .

Therefore $\sum{p_{i}} = 17 + 5 = \boxed{22}$ .