I started with bezout's theorem and assumed gcd(ax+by,ab) = d. From this I proved that d divides $a^(2)x$ & also $b^(2)y$. I intend to prove that d divides both a & b, which would be sufficient. If there's any alternative solution, please do suggest.
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Hint: assume some prime $p$ divides both $ax+by$ and $ab$. Then $p$ divides either $a$ or $b$. – Wojowu Apr 05 '19 at 15:48
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Suppose there is a prime $p\mid \gcd(ax + by, ab)$, then $p\mid ab$ and since $a,b$ are relatively prime $p$ must divide exactly one, say $a$.
Since $p\mid ax+by$ we conclude that $p\mid by$ and since $p$ does not devide $b$ we have $p\mid y$.
But then $p\mid \gcd(a,y)=1$, a contradiction!
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