2024 AIME I Problems/Problem 13: Difference between revisions

Revision as of 19:44, 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 15: / Line 15: @@
 {{AIME box|year=2024|n=I|num-b=12|num-a=14}}
+[[Category:Intermediate Number Theory Problems]]
 {{MAA Notice}}

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

Page

Toolbox

Search

2024 AIME I Problems/Problem 13: Difference between revisions

Revision as of 19:44, 2 February 2024

Problem

Solution

See also