Celebrating my comeback with "Primes" ^_^

By Fermat's theorem, $\forall a \in \mathbb{Z}$ and for all prime numbers $p$ ,

$a^p \equiv a \pmod{p}$

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This tells us

$17^p \equiv 17 \pmod{p}\\ 31^{p} \equiv 31 \pmod{p}\\19^{p} \equiv 19 \pmod{p}$

It follows that $31^{2p} \equiv 31^2 \equiv 961\pmod{p}\\ 19^{3p} \equiv 19^3 \equiv 6859 \pmod{p}$

Hence $17^p+31^{2p}+19^{3p} \equiv 17+961+6859 \equiv 7837 \pmod{p}$

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Given that $p\mid 17^p+31^{2p}+19^{3p}$

Hence $p \mid 7837$

And only prime factors of $7837$ are 17 and 461.

Answer $17+461 = \boxed{478}$