Competition problem — Number theory

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Number theory Difficulty 5.9 AIME, harder Prove it

Example 1 Let $a, b$ be positive integers, and $\frac{ab}{a+b}$ is also a positive integer. Prove: $(a, b)>1$ .

Solution

Prove that if $(a, b)=1$ , then $(a, a+b)=1$ (this can be deduced from property 3), hence, from $a+b \mid a b$ and $(a, a+b)=1$ , we get $a+b \mid b$ , but $a+b>b$ , so $a+b \mid b$ cannot hold. Therefore, $(a, b)>1$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.