Conjecture that $ \frac{\gcd(a+b,ab)}{\gcd(a,b)} \mid \gcd(a,b)$ I have discovered some exercise type conjectures which I can't prove and this is one of them:

Given positive integers $a,b$, then
  $$ \frac{\gcd(a+b,ab)}{\gcd(a,b)}\ \bigg|\ \gcd(a,b)$$

Can this be proved or disproved?

From time to time, when testing my growing math packages BigZ and Forthmath, I recognize some patterns which I can't prove or disprove (or even have the ambition to). I post them here with the hope that it will not annoy too much. I hope you can bear with it.

Write $\gcd(a,b)=d$, then $a=da',b=db'$ and thus $\frac{\gcd(a+b,ab)}{d}=\gcd(a'+b',a'b'd)$ where $\gcd(a',b')=1$. Notice now that $\gcd(a'+b',a'b')=1$ since $a'(a'+b')-a'b'={a'}^2$ and thus $\gcd(a'+b',a'b')|{a'}^2 \implies \gcd(a'+b',a'b')|a'$ and $\gcd(a'+b',a'b')|a'+b' \implies \gcd(a'+b',a'b')|a',b' \implies \gcd(a'+b',a'b')=1$. This means that $\gcd(a'+b',a'b'd)=\gcd(a'+b',d)$ and thus it divides $d$ by definition. So your conjecture is indeed true.
