Let $a, b$ and $c$ be integers. Prove that if $\gcd(a, b)$ and $\gcd(a, c)$ are coprime, then $\gcd(a, bc)$ = $\gcd(a, b) · \gcd(a, c)$
I am stumped in this problem. Can anybody clarify me what does $\gcd(a, b)$ and $\gcd(a, c)$ are coprime mean? Is that mean that $\gcd(a, b)$ = $\gcd(a, c) = 1$ ? Any hints on how to approach such problem? Thanks!
Let $x=\text{gcd}(a,b)$ and $y=\text{gcd}(a,c)$. We'll prove that $xy=\text{gcd}(a,bc)$ by showing $xy|\text{gcd}(a,bc)$ and $\text{gcd}(a,bc)|xy$.
First, $\text{gcd}(x,y)=1$ so there are integers $m,n$ such that $1=mx+ny$ which implies $$ a=m(xa)+n(ya). $$ Because $x|a$ and $y|a$, the RHS above is divisible by $xy$. This means the LHS $a$ is also divisible by $xy$. Moreover, $x|b$ and $y|c$ so $xy|bc$. Thus, $xy$ is a common divisor of $a$ and $bc$. In particular, $xy|\text{gcd}(a,bc)$.
Now, it remains to demonstrate that $\text{gcd}(a,bc)|xy$. For this, we can multiply the Bezout's identities for $x=\text{gcd}(a,b)$ and $y=\text{gcd}(a,c)$ to see that there are integers $r$ and $s$ satisfying: $$ xy=ar+(bc)s.\tag{A} $$ This implies $\text{gcd}(a,bc)|xy$, which completes our proof.
Edit: More details to get (A): write $$ x=ap+bq,\quad y=au+cv $$ and multiply to get $$ xy=a\underbrace{(apu+cpv+bqu)}_r+(bc)\underbrace{(qv)}_s. $$