# Thread: Show gcf(a,b) = gcf(a,c) = 1 => gcf(a,bc) = 1

1. ## Show gcf(a,b) = gcf(a,c) = 1 => gcf(a,bc) = 1

(i) Let a, b, and c be integers with gcf(a,b) = gcf(a,c) = 1. Show that gcf(a,bc) = 1.

There's a more general case to this in (ii) of the same problem, where if $${a_i}$$ is relatively prime with b for $1 \le i \le n$ then the product of $${a_i}$$ is relatively prime with b.

If you could give me a hint on how to prove (i), that would be greatly appreciated.

(i) Let a, b, and c be integers with gcf(a,b) = gcf(a,c) = 1. Show that gcf(a,bc) = 1.

There's a more general case to this in (ii) of the same problem, where if $${a_i}$$ is relatively prime with b for $1 \le i \le n$ then the product of $${a_i}$$ is relatively prime with b.

If you could give me a hint on how to prove (i), that would be greatly appreciated.
Since $gcd(a,b)=1$ and $gcd(a,c)=1$, there exist integers $s, t, u,$ and $v$ such that $as+bt=1$ and $au+cv=1$. So $(as+bt)(au+cv)=a(su+csv+btu)+bc(tv)=1$.

The general case: Suppose that $gcd(a_i,b)=1$, where $1\le i\le n$, then $gcd(a_1a_2\cdots a_n,b)=1$.
It's a corrolary of problem (i) and can be proved using induction on $n$; the case $n=2$ is the main result.

3. Is it usually the case that for gcf proofs, you recourse back to the integral sums? Thanks again for your help!

4. can someone show how the induction would go for part (2) in this question?