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The problem is the following:

Let $a$ and $b$ be non-zero integers. Let there be two integers $s$ and $t$ such that $sa + tb = 6$.

What can be said about the common divisors of $a$ and $b$ based on this equality?

If someone could help me out with this problem, I would appreciate it very much. I have tried solving it and have made no viable progress. My guess is that the problem is somewhat simple and most importantly easy, but I am missing that starting point.

Luka Duranovic
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    You can say that any common divisor $d$ divides $6$, but not much more. This is quite restrictive though, because it means that the (positive) common divisors of $a$ and $b$ comprise at most $1$, $2$, $3$, and $6$. – pancini Nov 07 '19 at 02:15

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Welll if $d|a$ then $d|sa$ and if $d|b$ then $d|tb$. And anything that divides $sa$ and also divides $tb$ will divide the sum of $sa$ and $tb$ so......

fleablood
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  • Do you think that that is all we can say for this? By the way thanks for the answer – Luka Duranovic Nov 07 '19 at 02:35
  • @Luka Yes, that's all we can say because it is solvable for $,s,t \iff \gcd(a,b)\mid 6\ \ $ (hint: scale the Bezout identity for $,\gcd(a,b)).\ $ This might be an exercise providing motivation for that general result. – Bill Dubuque Nov 07 '19 at 02:47
  • Well, I didn't actually spell out the conclusion that any common divisor must divide $6$. But that's pretty much it. – fleablood Nov 07 '19 at 05:48
  • I suppose it's "more" to say that the only common (positive) divisors are $1,2,3$ or $6$. That follows for saying any common divisor divides $6$ as only $1,2,3$ or $6$ divide $6$. And it need not be that $2,3$ or $6$ are common divisors. It could be that that only $1$ is a common divisor (if $a= 7$ and $b=4$ and $s=-6$ and $t=12$ then $sa+tb = 6$) or that the $\gcd(a,b)=2$ or $3$. – fleablood Nov 07 '19 at 06:04