$(a,b)\!=\!1\!=\!(a,c)\Rightarrow (a,bc)\!=\!1$ [coprimes to $\,a\,$ are closed under products] 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. 
 A: 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. 
A: More generally $\ \bbox[5px,border:1px solid #c00]{(a,b)(a,c) = (a,bc)}\ $ if $\  \color{#0a0}{(a,b,c)=1},\, $ follows by expanding the product, i.e. $\ (a,b)(a,c) = (aa,ab,ac,bc) = (a\color{#0a0}{(a,b,c)},bc) = (a,bc),\, $  using GCD polynomial arithmetic, i.e. the GCD associative, commutative, and distributive laws (explained in the prior link).
OP is special case $\ (a,b)\! =\! \color{#c00}{\bf 1}\! =\! (a,c),\ $ so $\,\color{#0a0}{(a,b,c)\! =\! 1},\ $ so $\  \color{#c00}{\bf 1\cdot 1} = (a,bc)\ $ by above.
Remark $ $ Worth emphasis: the common Bezout-based proofs can be viewed more conceptually as: invertibles are closed under product, via $\,b,c\,$ are coprime to $\,a\!\iff\! b,c\,$ invertible $\!\bmod a,\,$ by Bezout. The proof is obvious: $\,\overset{}{bb'}\equiv 1\equiv cc'\Rightarrow\ bc\,\color{#c00}{c'b'}\equiv 1\, $ (i.e. $\,(bc)^{-1}\equiv \color{#c00}{c^{-1} b^{-1}})$.
In fraction language it's obvious $\,1/(bc) \equiv (1/b)(1/c).\,$ So the proof becomes obvious and intuitive after translating the Bezout equations into more arithmetical congruence language - being essentially the same as well known inverse or fraction laws.
The above proof by GCD laws is more general than proofs using Bezout's linear gcd reprentation, since generally gcds fails to have this form, e.g. in the UFD $\rm\,D = \mathbb Q[x,y]\,$ of polynomials in $\rm\,x,y\,$ with rational coefficients we have $\rm\,gcd(x,y) = 1\,$ but there are no $\rm\,f(x,y),\, g(x,y)\in D\,$ such that $\rm\,x\,f(x,y) + y\,g(x,y) = 1;\,$ indeed, if so, then evaluating the purported Bezout equation at $\rm\:x = 0 = y\:$ yields $\,0 = 1.\,$ Similarly for $\,\gcd(2,x)=1\,$ in the UFD $\rm\,\Bbb Z[x]$.
Further, such GCD arithmetic is simpler than Bezout equation manipulations since it is essentially the same as (high-school) polynomial arithmetic  - as explained in the first link above.
A: 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.
A: Since $a, b$ are relatively prime, there exist integers $x, y$ such that $ax + by = 1$. Multiplying both sides of this equation by $c$, we get $$acx + bcy = c.$$ Letting $d = \gcd(a, bc)$, we see from the equation above that since $d$ is a common divisor of $a$ and $bc$, $d$ also divides $c$ by linearity. Then, since every common divisor of two integers also divides their gcd, $d|\gcd(a, c)$. But $\gcd(a, c) = 1$, so $d$ divides $1$. Because the gcd is nonnegative, it follows that $d = \gcd(a, bc) = 1$. 
A: 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, $(\color{#0000FF}{a},\color{#0000FF}{bc})=1$
A: *

*$\gcd(a,b)=1$ means what about the prime factors common to both $a,b$?

*same for $\gcd(a,c)$

*what is the prime decomposition of $bc$ if you know what $b$, $c$ decompose as?


Combine all together...
A: $\gcd(a,b)$ means a and b are relatively prime, this means, there is no other common factor other than 1.
you also know that $$\gcd(a,b) = 2^{\min(a_1,b_1)}.3^{\min(a_2,b_2)}.5^{\min(a_3,b_3)}...$$
Now the 2 equations already imply that $$\min(a_1,b_1) = \min(a_2,b_2)=...= 0$$ (there is no common factor of primes)
also $$\min(a_1,c_1) =\min(a_2,c_2)=... = 0$$ 
this imply, $$\min(a_1,b_1+c_1)= \min(a_2,b_2+c_2)=...=0$$ as the powers > 0, and adding them cannot be < 0, 
This imply a,bc are also relatively prime, hence $$\gcd(a,bc)=2^{\min(a_1,b_1+c_1)}.3^{\min(a_2,b_2+c_2)}...= 1$$
($\min(a,b)$ represents here minimum value of a,b)
