proofs of gcd. i am very confused how to solve it

Question

How do I go about proving this?

If $gcd(a,b)=1$ and $gcd(a,c)=1$, then $gcd(a,bc)=1$. I'm very confused with gcd proofs.

Answer 1

If $gcd(a, b) = gcd(a, c) = 1$ then there exist $m, n$ such that $am + bn = 1$ and $s, t$ such that $as + ct = 1$.

Multiplying the first equality by $ct$ gives $amct + bnct = ct$, so $as + amct + bnct = as + ct$ and so $a(s + mct) + bc(nt) = 1$, which implies that $1 = gcd(a, bc)$

Or

There are $x$ and $y$ such that $bx ≡ cy ≡ 1 ($mod $a)$, so $(yx)bc ≡ y(xb)c ≡ yc ≡ 1 ($mod $a)$. Since $bc$ has a multiplicative inverse mod $a, gcd(a, bc) = 1$.

Answer 2

Let $a = p_{1}^{\alpha_{1}}\dots p_{k}^{\alpha_{k}}$ and $b=q_{1}^{\beta_{1}}\dots q_{l}^{\beta_{l}}$, and $c = r_{1}^{\gamma_{1}}\dots r_{m}^{\gamma_{m}}$.

Using gcd we have that there is no $p_{i} = q_{j}$ for all $i,j$ also for $p_{i}$ and $r_{j}$. So there is no equal $p_{i}$ and $r_{j} q_{y}$ in $bc = q_{1}^{\beta_{1}}\dots q_{l}^{\beta_{l}}r_{1}^{\gamma_{1}}\dots r_{m}^{\gamma_{m}}$