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2023 SSMO Team Round Problems/Problem 4

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Problem

Find the sum of values for prime $p$ such that $p \mid (2023^{p^2}+(p-1)!+2^{p^4}).$

Solution

We have that $\varphi(p) = p - 1$, so by Fermat’s Little Theorem, \[a^p \equiv a \pmod{p}.\]

Thus, \[2023^{p^2} + 2^{p^4} \equiv 2023 + 2 \equiv 2024 \pmod{p}.\]

In other words, $p \mid 2024$. The prime divisors of 2024 are $2$, $11$, and $23$, so the answer is \[2 + 11 + 23 = \boxed{36}.\]

~SMO_Team

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