$\gcd(a,b)\!=\!1\!=\!\gcd(a,c)\Rightarrow\gcd(a,bc)\!=\!1$ [coprimes to $\,a\,$ are product closed]

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.

Solution 1:

Using Bezout's Identity:

Since there are $x,y,u,v$ so that $\color{#C00000}{ax+by=1}$ and $\color{#00A000}{au+cv=1}$, we have $$ \begin{align} \color{#C00000}{by}\color{#00A000}{cv} &=\color{#C00000}{(1-ax)}\color{#00A000}{(1-au)}\\ &=1-a(x+u-axu)\\ \color{#0000FF}{a}(x+u-axu)+\color{#0000FF}{bc}vy &=1 \end{align} $$ Therefore, $(\color{#0000FF}{a},\color{#0000FF}{bc})=1$

Solution 2:

If you know Euclid's Lemma (if $p$ is a prime number and $p$ divides the product $bc$, then either $p$ divides $b$ or $p$ divides $c$), then you can proceed as follows. Let $d=gcd(a,bc)$ and assume, to arrive at a contradiction, that $d>1$. Let $p$ then be a prime number dividing $d$. Then $p$ divides both $a$ and $bc$, and thus either $p$ divides $b$ or it divides $c$. If $p$ divides $b$, then since it also divides $a$, it follows that $p$ divides $gcd(a,b)=1$, which is absurd. Similarly, $p$ dividing $c$ leads to a contradiction.

Remark: this solution avoids considering the entire prime number decomposition for the given numbers. I find it to be a nice little trick that produces clean proofs.

Solution 3:

We can take as the definition of "$a,b$ are relatively prime" that there are $s,t$ with $1=as+bt$. Then the question becomes: given that$$1=as+bt$$and$$1=as^\prime+ct^\prime$$find $S,T$ with$$1=aS+(bc)T.$$Multiplying by $1$ is a useful trick/technique for questions of this sort.$$1=as+bt=as+\mathbf{1}bt=as+\mathbf{(as^\prime+ct^\prime)}bt$$so$$\begin{align}1&=a(s+cst^\prime)+bc(tt^\prime)\\&=a(s^\prime+bs^\prime t)+bc(tt^\prime).\end{align}$$

The second formula arises from the first by applying $(bc)(ss^\prime)(tt^\prime)$ or shifting the derivation. A single formula invariant under that permutation is

$$1=(as+bt)(as^\prime+ct^\prime)=a(ass^\prime+bts^\prime+cst^\prime)+bc(tt^\prime).$$

The symmetry is nice but the first formulas are smaller.

What I like is that uses nothing more than basic properties of rings so the proof is applicable in many rings.