Number Theory GCD Homework Problem

This is a problem from my homework:

If $a\mid bc$, show that $a\mid$gcd$(a,b)$gcd$(a,c)$.

(Hint: use Euclid's lemma: If $a_0\mid b_0c_0$, with gcd$(a_0,b_0)=1$, then $a_0\mid c_0$.)

I tried setting $a_0=a$, $c_0=c$, and $b_0=$gcd$(a,b)$, to try to find gcd$($gcd$(a,b),a)$, but I stopped here as I wasn't able to go further.

Thanks in advance!

Answer

$a_0=a,b_0=\gcd(a,b)$ does not meet the condition $\gcd(a_0,b_0)=1$

The key is that $$\gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)=1$$

let $$a_0=\frac{a}{\gcd(a,b)},b_0=\frac{b}{\gcd(a,b)},c_0=c$$
we have $$\begin{align}a|bc\Rightarrow & a_0*\gcd(a,b)|b_0*\gcd(a,b)*c_0\\\Rightarrow & a_0|b_0c_0\end{align}$$
and $$\gcd(a_0,b_0)=1$$ Thus $$a_0|c_0 \Rightarrow \frac{a}{gcd(a,b)}|c$$

obviously $$\frac{a}{gcd(a,b)}|a$$ . So $$\frac{a}{gcd(a,b)}|\gcd(a,c)$$
ie. $$a|\gcd(a,b)\gcd(a,c)$$