Competition problem — Number theory

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Number theory Difficulty 4.6 AIME Prove it Soviet Union

Problem:
If $m$ , $k$ , $n$ are natural numbers and $n > 1$ , prove that we cannot have $m(m + 1) = k^n$ .

Solution

Solution:
$m$ and $m + 1$ have no common divisors, so each must separately be an $n$ th power. But the difference between the two $n$ th powers is greater than $1$ (for $n > 1$ ).

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