Show that $ax \equiv b (mod\ m) $ has solution iff $gcd(a,m)$ divides $b$

Question

Here's what I have:

$ax \equiv b (mod\ m)$ has answer if there are $x$ and $y$ such that

$b = ax + my$

Let $d = gcd(a,m)$. Then:

Since $m$ divides the right part of the equation, it also has to divide the left part.

Is this a valid proof for what I want?

You have half of it: "If there is a solution, then d divides b." You also need to show that whenever b is a multiple of d, you have a solution. — Chessanator, Jun 12 '17 at 17:07
Have you seen yet that the gcd of a and m is a linear combination of them? — Chessanator, Jun 12 '17 at 17:24
Yes, that's what DonAntonio used bellow, I understand. Thanks! — dumb_undergrad, Jun 12 '17 at 17:34

score 1 · Accepted Answer · answered Jun 12 '17 at 17:25

1

If $\;d=gcd(a,m)\;$, then there exist $\;r,s\in\Bbb Z\;$ s.t. $\;ra+sm=d\;$ , so

$$b=cd\implies b=c(ra+sm)=a(cr)+m(cs)$$

and we have a solution.

answered Jun 12 '17 at 17:25

DonAntonio

That completes my proof, as Chessnator said above, I've done only half of it. Now, how does that show we have a solution? – dumb_undergrad Jun 12 '17 at 17:35
@dumb_undergrad I don't understand: in the above it is clear that $;x=cr,,y=cs;$ is a solution... – DonAntonio Jun 12 '17 at 19:36
Oh, right, thanks! – dumb_undergrad Jun 12 '17 at 19:43

1 Answers1