Difference between revisions of "2024 AIME I Problems/Problem 13"

Revision as of 18:44, 2 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

@@ Line 4: / Line 4: @@
 ==Solution==
+<math>n^4+1\equiv 0\pmod{p^2}\implies n^8 \equiv 1\pmod{p^2}\implies p_{min}=17</math>
+From there, we could get <math>n\equiv \pm 2, \pm 8\pmod{17}</math>
+By doing binomial expansion bash, the four smallest <math>n</math> in this case are <math>110, 134, 155, 179</math>, yielding <math>\boxed{110}</math>
+~Bluesoul
 ==See also==

2024 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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Difference between revisions of "2024 AIME I Problems/Problem 13"

Revision as of 18:44, 2 February 2024

Problem

Solution

See also