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Let $a,b,c$ be three integers whose greatest common divisor is $1$ (ie $\gcd(a,b,c)=1$). Show that there exist integers $m$ and $n$ such that $a+mc$ and $b+nc$ are coprime.

Progress: I believe the question is essentially asking us to show that it's always possible to find integers $p$ and $q$ and $s$ such that $pa+qb+sc=1$, and $p$ and $q$ are coprime.

Added Context

This related question indicates that this problem requiring $n=0$, is more complex. This question, by not requiring $n=0$, might allow simpler solution.

Community
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OrangeApple3
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  • Yes. Thank you. – OrangeApple3 Apr 11 '15 at 04:14
  • It may also be good to write explicitly $\gcd(a,b,c)=1$, so in particular $\gcd(a,b)$ could be anything. – Marc van Leeuwen Apr 11 '15 at 04:15
  • This related question indicates that the related problem, requiring $n=0$, is more complex. This question is related, but allows a far simpler solution by not requiring $n=0$. This is an interesting question and I think it should be reopened. – robjohn Apr 11 '15 at 14:08
  • Technically, my old question isn't that hard, it was just using the condition of avoiding a unique factorization proof that made it hard. The old question does appear to require some sort of descent argument, however. @robjohn – Thomas Andrews Apr 11 '15 at 14:39

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Bezout says there are $x,y$ and $u,v$ so that $ax+by=(a,b)$ and $(a,b)u+cv=1$. Thus, $$ \begin{align} &(a+yc)\left(ux+\frac{bv}{(a,b)}\right)+(b-xc)\left(uy-\frac{av}{(a,b)}\right)\\ &=\color{#600000}{a\left(ux+\frac{bv}{(a,b)}\right)+b\left(uy-\frac{av}{(a,b)}\right)}\\ &+\color{#005000}{yc\left(ux+\frac{bv}{(a,b)}\right)-xc\left(uy-\frac{av}{(a,b)}\right)}\\[6pt] &=\color{#600000}{(a,b)u}+\color{#005000}{cv}\\[12pt] &=1 \end{align} $$ Therefore, $$ (a+yc,b-xc)=1 $$

robjohn
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  • Nice direct Bezout proof. You can just write it as $(a+yc)b-a(b-xc)=c(a,b)$. Dividing by $(a,b)$, we see that the gcd of $a+yc$ and $b-xc$ is a divisor of $c$. But if $p$ divides $c$ and $a+yc$ and $b-xc$ then $p\mid (a,b,c)$. – Thomas Andrews Apr 11 '15 at 14:47