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let $a,b,c\in Z$ and $(x,y) \in Z$ is a solution of $ax+by=c$, prove that $c$ is divisible by $\gcd(a,b)$.

All hints are welcome, I need a point where I can start from.

Clyde Kertzer
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SuperNova
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    Both $ax$ and $by$ are divisible by it. – Vladimir Jun 09 '14 at 13:41
  • thanks, how can I write that in correct formal speech? ax|gcd(a,b) and by|gcd(a,b) so ax+by must be | gcd(a,b) – SuperNova Jun 09 '14 at 13:46
  • I do not know how you can do that; I did so in my comment. – Vladimir Jun 09 '14 at 13:46
  • Yes, you did good – Vladimir Jun 09 '14 at 13:48
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    You can write that $\gcd(a,b)$ divides $ax$. If you use $\mid$, this is written $\gcd(a,b)\mid ax$. – André Nicolas Jun 09 '14 at 14:24
  • Your claim is not true for $x,y\in\mathbb R$. However, it is true for $x,y\in\mathbb Z$. This is likely what you meant and so it should be added to the question. – user26486 Jun 09 '14 at 15:26
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    @SuperNova By definition the gcd of $,a,b,$ is a common divisor of $,a,b,,$ i.e. $,d = \gcd(a,b)\mid a,b,,$ so $,d\mid ax,by,,$ so it divides their sum. In other words, the multiples of $,d,$ are closed under addition, and by multiplication by arbitrary integers (i.e. they form an ideal). – Bill Dubuque Jun 09 '14 at 17:02

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As a few commentors have pointed out, this proof is quite straightfoward. Clearly, $\gcd(a,b) \mid ax$ and $\gcd(a,b) \mid by$. Thus, $\gcd(a,b) \mid (ax+by)$. Because $ax + by = c$, we also know that $\gcd(a,b) \mid c$, and we are done.

Clyde Kertzer
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