I've been doodling around with some numbers and I think this is true, however I have not been able to prove it yet. I'd like to know if you have a suggestion or a counterexample for this. Thanks!
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3Choose $a=d, b=1$ – player3236 Nov 11 '20 at 04:55
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If $m=12$ and $n=18$, then $d=6$. If I choose $a=d, b=1$, then $1\neq 2=(2, 18)=(\frac{m}{a}, \frac{n}{b})$. Moreover, if I choose $a=1, b=d$, then $1\neq 3=(12, 3)=(\frac{m}{a}, \frac{n}{b})$. – Bruno Aceves Nov 11 '20 at 05:05
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My oversight. In that case, one could consider the prime factorization of $m,n$ and choose $a,b$ to according to the exponent of each prime in the factorization. – player3236 Nov 11 '20 at 05:09
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Ok, I will try it, thanks for the suggestion. :) – Bruno Aceves Nov 11 '20 at 05:20