Competition problem — Number theory

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Number theory Difficulty 4.8 AIME Prove it Soviet Union

Problem:

The natural numbers $m$ and $n$ are relatively prime. Prove that the greatest common divisor of $m + n$ and $m^2 + n^2$ is either $1$ or $2$ .

Solution

Solution:

If $d$ divides $m + n$ and $m^2 + n^2$ , then it also divides $(m + n)^2 - (m^2 + n^2) = 2mn$ and hence also $2m(m + n) - 2mn = 2m^2$ and $2n(m + n) - 2mn = 2n^2$ . But $m$ and $n$ are relatively prime, so $m^2$ and $n^2$ are also. Hence $d$ must divide $2$ .

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