If $\gcd(a,b)= 1$ and $a$ divides $bc$ then $a$ divides $c\ $ [Euclid's Lemma] Well I thought this is obvious. since $\gcd(a,b)=1$,  then we have that $a\not\mid b$ AND $a\mid bc$. This implies that $a$ divides $c$. But apparently this is wrong. Help explain why this way is wrong please. 
The question tells you give me two relatively prime numbers $a$ and $b$ such that $a$ divides the product $bc$, prove that $a$ divides $c$. how is this not obvious? Explain to me how my "proof" is not correct.

Since $\gcd(a,b) = 1$, you can choose integers $x$ and $y$ so that $ax + by = 1$.  Hence, $axc + byc = c$.  Suppose $a| bc$; write $aq = bc$. Then $c = axc + yaq$, so $a|c$.  
