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Let $a,b,c,d,x,y \in \mathbb{Z}$ such that $\gcd(a,b) = 1, \gcd(a,c) = 1, \gcd(d,b) = 1, \gcd(d,c) = 1, \gcd(x,y) = 1$.

Prove $\gcd(ax+by,cx+dy) = 1$

I'm trying to figure out how I can use these facts that parts of this formula $ax+by$ and $cx+dy$ are coprime to say that their linear combination must also be coprime. But I'm not sure how to formalize this since I'm dealing with sums rather than products.

ENV
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    This is true instead https://math.stackexchange.com/q/1784008/399263, see jjagmath counterexample. – zwim Oct 13 '21 at 16:14
  • Thanks very much @zwim! That's actually exactly what I was looking for – ENV Oct 13 '21 at 16:22

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Counterexample: $a=b=c=d=x=y=1$

jjagmath
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