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Harvard-MIT Mathematics Tournament

Number theory Difficulty 4.6 AIME Prove it United States

Problem:
Find all positive integer solutions (m,n)(m, n) to the following equation:
m2=1!+2!+⋯+n!. m^{2} = 1! + 2! + \cdots + n!.

Solution

Solution:
(1,1),(3,3)(1,1), (3,3)
A square must end in the digit 0,1,4,5,60, 1, 4, 5, 6, or 99. If n≥4n \geq 4, then 1!+2!+⋯+n!1! + 2! + \cdots + n! ends in the digit 33, so cannot be a square. A simple check for the remaining cases reveals that the only solutions are (1,1)(1,1) and (3,3)(3,3).

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.