If gcd(a, b) = 1, Then gcd(a+b, ab) = 1:

Assume \(\gcd(a+b, ab) = d\), then there is a prime \(p\) that divides \(d\). If \(p|ab\), then \(p|a\) or \(p|b\) (click here for proof).

Assume \(p|a\). If \(p|a\) and \(p|(a+b)\), then \(p|b\).

If \(p\) divides both \(a\) and \(b\), then we have a contradiction (since \(\gcd(a, b)=1\)).

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