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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 $\gcd (a, b)$.

Darth Geek
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  • Contradiction is unnecessary:

    $\gcd(a,b)$ divides $a$ and $b$. Therefore it divides $ax+by$ for any integers $x,y$. Ergo if $ax+by = c$ it follows that $\gcd(a,b)$ divides $c$ as well.

     – Darth Geek Aug 13 '14 at 21:04
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    You probably mean that $\gcd(a,b)$ divides $c$, not vice versa. – Andreas Blass Aug 13 '14 at 21:04
  • Could you please explain why is it necessary for the solutions to be integers? – Charith May 16 '17 at 15:05

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You can prove it like that:

Let $d=(a,b)$

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

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

evinda
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