Prove that if $\gcd(a,b)=1$ then $\gcd(a,bc) = \gcd(a,c)$.

Question

Prove that if $\gcd(a,b)=1$ then $\gcd(a,bc) = \gcd(a,c)$.

Found many similar questions to this, but none the same. I've been trying to use Bezout's Lemma, but haven't figured it out.

Answer 1

One can argue by the Fundamental Theorem of Arithmetic.

Let $a=\prod_{i=1}^{n}p_i^{\alpha_i}$ be the prime factorisation of $a$. Since $\gcd(a, b)=1$, the primes in the prime factorisation $b=\prod_{j=1}^{m}{p'}_j^{\beta_j}$ of $b$ are distinct from each of those in that of $a$.

Consider, then, the prime factorisation $c=\prod_{r=1}^{t}{p''}_r^{\gamma_r}$.

By the Fundamental Theorem of Arithmetic, the factorisation of $bc$ is $$bc=\prod_{j=1}^{m}{p'}_j^{\beta_j}\prod_{r=1}^t{p''}_r^{\gamma_r}.$$

Can you continue from here?

Answer 2

Lemma: In $\mathbb{Z}$, if $a,b>0$ and $a\mid b$ and $b\mid a$ then $a=b$.

Let $t=\gcd(a,bc)$ and $s=\gcd(a,c)$.

Since $t=\gcd(a,bc)$ this means:

1. $t\mid a$

2. $t\mid bc$

3. If $r\mid a$ and $r\mid bc$ then $r\mid t$.

The goal will be to show that $t\mid s$ and $s\mid t$ and then by the lemma, $t=s$.

To show that $s\mid t$, it suffices to show that $s\mid a$ and $s\mid bc$.

Hint: Show $s$ and $b$ are coprime.

Answer 3

According to Bezout the gcd of $a$ and $b$ is the smallest positive integer combination of $a$ and $b$. Thus there exist integers $m$ and $n$ satisfying $$\mathrm{gcd}(a,bc) = am + bcn = am + c(bn) \ge \mathrm{gcd}(a,c).$$ Likewise, there exist integers $x$ and $y$ satisfying $ax + by = 1$ and integers $p$ and $q$ satisfying $$\mathrm{gcd}(a,c) = ap + cq = (ap + cq)(ax + by) = a(pax + cqx + bpy) + bc(y) \ge \mathrm{gcd}(a,bc).$$

Answer 4

To use Bezout.

You have $x,y, w,u$ so that

$ax + by = 1$ and $aw + cu = \gcd(a,c)$.

The trick is to replace $1$ with $ax+by$ and $u$ with $u(ax + by)$

So $aw + c(u(ax+by)) =$

$a(w+cux) + bc(uy) =\gcd(a,c)$.

So $\gcd(a,bc)|\gcd(a,c)$.

Now $\gcd(a,c)|a$ and $\gcd(a,c)|c$ so $\gcd(a,c)|bc$. So $\gcd(a,c)|\gcd(a,bc)$.

So $\gcd(a,bc) = \gcd(a,c)$.