TITLE: Check $\,\gcd(a,bc)=\gcd(a,\gcd(a,b)c)\,$ for integers $a,b,c$
QUESTION [5 upvotes]: For the integers $a,b,c$ show that $\gcd(a,bc)=\gcd(a,\gcd(a,b)\cdot c)$
Proof:
Let $u$ and $v$ be integers. Then $\gcd(a,b)=au+bv$. Then $c\cdot \gcd(a,b)=c\cdot(au+bv)=acu+bcv$
Let $x$ and $y$ be integers. Then $\gcd(a,\gcd(a,b)\cdot c)=ax+(acu+bcv)y=ax+acuy+bcvy=a(x+cuy)+(bc)(vy)$.
Let $d$ and $e$ be integers. Then $\gcd(a,bc)=ad+bce$. 
Since $d$ and $e$ are integers let $d=x+cuy$ and $e=vy$. Hence $\gcd(a,bc)=\gcd(a,\gcd(a,b)\cdot c)$

REPLY [2 votes]: Since a proof has already been sketched by Bill, let me give you an alternative solution/approach to the problem that you mind find intuitive.
First off, look at the formula that you're asked to establish:
$$
    \gcd(a,bc) = \gcd(a,\gcd(a,b)\cdot c).
$$
Can you translate this equation into a simple English sentence? Really think about what it means given the meaning of GCD. Here's my attempt, which is actually a slightly stronger statement:

 A divisor of $a$ also divides $bc$ if and only if part of it divides $b$ and the rest of it divides $c$.

Thinking about it in these terms suggests an elementary proof. (I'll let you fill in any small gaps below.) Let $d = \gcd(a,b)$. Then $a = da'$ and $b = db'$ for relatively prime integers $a',b'$. Thus $\gcd(a,bc) = \gcd(da',db'c) = d\gcd(a',b'c)$. Now to say that $a'$ and $b'$ are relatively prime means that they share no common prime factors. So a prime $p$ divides both $a'$ and $b'c$ if and only if it divides $a'$ and $c$. As a consequence $\gcd(a',b'c) = \gcd(a',c)$, and therefore $D = \gcd(a,bc) = d\gcd(a',b'c) = d\gcd(a',c) = \gcd(da',dc) = \gcd(a,dc)$ as claimed.