elementary number theory - $gcd (ca, cb) = gcd (a, b)c$ if $c > 0$

Let $\gcd (a, b) = d$. So, $ax + by = d$ for some $x, y$. Then $(ca)x + (cb)y = cd$. Thus, $\gcd (ca, cb) = cd = \gcd(a, b)c$.

Does it work?

Answer

No. The problem is that $ax+by = d$ does not imply $\gcd(a,b) = d$.

The statement $\gcd(ca,cb) = c\gcd(a,b)$ is true, however. To see that note that

$\gcd(a,b) \mid a \wedge \gcd(a,b) \mid b \implies c\gcd(a,b) \mid ac \wedge c\gcd(a,b) \mid bc$ so
$$\gcd(ac,bc) \mid c\gcd(a,b)$$

The reverse is a little bit more work (you need $\gcd(ac,bc) \mid c\gcd(a,b)$ to conclude).