How to prove the following statement: $\gcd(a,b) = 1$ and $a=bc$ implies $a\mid c$

Question

If $a$ and $b$ are coprime, i.e. $\gcd(a,b) = 1 = ax+by$, for certain integer $x$ and $y$, and $a = bc$, then $a \mid c$

I've tried multiply $c$ to both sides of $1 = ax+by$, but didn't work.

Answer 1

As $a=bc$, $a|bc$.

From Bezout's identity, there exist $x,y$ such that $xa+yb=1$. Multiply both sides by $c$ and get $$xac+ybc=c.$$ $xac$ is divisible by $a$, $ybc$ is divisible by $bc$, which is divisible (by hypothesis) by $a$. So $c$ must also be divisble by $a$.

Answer 2

Still it should work if you use all the hypotheses: you obtain $c=acx+bcy$. Now, $a$ is supposed to divide $bc$, i.e. there exists $k\in \mathbf Z$ such that $bc=ak$, so $$c=acx+aky=a(cx+ky).$$