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I've been stuck on this one for awhile, and I thought that if I did a contradiction: if $\gcd(m,n) > 1$ then for all integers $x,y$, $ax + ny \neq 1$. I could simply prove it wrong by giving an example. But I've got a feeling that I'm missing something that makes this not work, But I don't know what else to do.

Bill Dubuque
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AY99
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    if I did a contradiction But what you wrote next is the wrong premise for a proof by contradiction. Instead, you'd start with something like "suppose $\gcd(m,n) \ge 2$ and $mx+ny=1$ then...". – dxiv Jan 16 '18 at 03:53
  • The negation of $p \implies q$ is "p and not q". The statement you wrote is "(not p) implies q" – Air Conditioner Jan 16 '18 at 03:57
  • Please, use MathJax to format mathematical expressions. – mucciolo Jan 16 '18 at 04:06
  • By the linked dupe, since $,m,n,$ are multiples of the gcd so too is their integral linear combination $,mx+ny = 1,,$ so the gcd $= 1.\ \ $ – Bill Dubuque Jul 28 '22 at 12:45

2 Answers2

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As the comments noted, the contradiction assumptions in the original question are not correct. The correct assumption (and some hints of how to proceed) follow:

Suppose that $\gcd(m,n)=d>1$ and suppose that there exist integers $x$ and $y$ so that $mx+ny=1$. Since $d$ divides both $m$ and $n$ (since it is the greatest common divisor of them), $d$ divides the LHS. So $d$ divides the RHS. Now, can you find a contradiction?

Michael Burr
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$\mathrm{gcd}(m,n)$ divides $mx+ny$ for all $x,y\in\Bbb{Z}$, and only $1$ divides $1$.

QED
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