prove by contradiction that $ax+by=c$ has no integer solutions if $c$ does not divide into $\gcd (a, b)$

Question

Prove by contradiction that (the diophantine equation) $ax+by=c$ has no integer solutions if $c$ does not divide into $\gcd (a, b)$. Here is what I did: lets assume $c$ divides into $\gcd (a, b)$. There are infinitly many solutions if $c$ divides $\gcd (a, b)$.

Answer 1

You can prove it like that:

Let $d=(a,b)$

If $ax+by=c$ has a solution $x_1,y_1$,then:

$d \mid a, d \mid b \Rightarrow d \mid ax_1+by_1=c$