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Difference between revisions of "2023 SSMO Team Round Problems/Problem 4"

(Created page with "==Problem== Find the sum of values for prime <math>p</math> such that <math>p \mid (2023^{p^2}+(p-1)!+2^{p^4}).</math> ==Solution==")
 
 
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==Solution==
 
==Solution==
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We have that <math>\varphi(p) = p - 1</math>, so by Fermat’s Little Theorem,
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<cmath>a^p \equiv a \pmod{p}.</cmath>
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Thus,
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<cmath>2023^{p^2} + 2^{p^4} \equiv 2023 + 2 \equiv 2024 \pmod{p}.</cmath>
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In other words, <math>p \mid 2024</math>. The prime divisors of 2024 are <math>2</math>, <math>11</math>, and <math>23</math>, so the answer is
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<cmath>2 + 11 + 23 = \boxed{36}.</cmath>
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~SMO_Team

Latest revision as of 21:17, 9 September 2025

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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