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Difference between revisions of "2024 AIME I Problems/Problem 13"
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~Bluesoul
 
~Bluesoul
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==Video Solution 1 by OmegaLearn.org==
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https://youtu.be/UyoCHBeII6g
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==See also==
 
==See also==

Revision as of 13:54, 3 February 2024

Problem

Let $p$ be the least prime number for which there exists a positive integer $n$ such that $n^{4}+1$ is divisible by $p^{2}$. Find the least positive integer $m$ such that $m^{4}+1$ is divisible by $p^{2}$.

Solution

$n^4+1\equiv 0\pmod{p^2}\implies n^8 \equiv 1\pmod{p^2}\implies p_{min}=17$

From there, we could get $n\equiv \pm 2, \pm 8\pmod{17}$

By doing binomial expansion bash, the four smallest $n$ in this case are $110, 134, 155, 179$, yielding $\boxed{110}$

~Bluesoul

Video Solution 1 by OmegaLearn.org

https://youtu.be/UyoCHBeII6g


See also

2024 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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