Let $a$, $b$, $c$ be integers. Show that if $a$ and $b$ are coprime and $a$ and $c$ are coprime then $a$ and $bc$ are coprime.
My approach:
Suppose that $a$ and $b$ are coprime, $a$ and $c$ are coprime but $a$ and $bc$ are not coprime. This implies $gcd(a,b)=1$, $gcd(a,c)=1$ and that $\exists$ $d$ $\in$ $\Bbb N$ such that $gcd(a,bc)=d$ where $d>1$. Since $d>1$ then $\exists$ $p$ a prime number such that $p|d$.
We know that $p|d$ and $d|a$, this implies $p|a$. Also, $p|d$ and $d|bc$, this implies $p|bc$. By Euclid's lemma, we have two cases:
Case 1: $p|a$ and $p|b$, this implies $gcd(a,b)\ge p$. Since $p\ne1$, we have contradiction.
Case 2: $p|a$ and $p|c$, this implies $gcd(a,c)\ge p$. Since $p\ne1$, we have contradiction.
Since these cases are collectively exhaustive, then we have contradiction and $a$ and $bc$ must be coprime.
Is this approach sound? Is there an easier way?
