Is it true that $(a,b)=(c,b)=1\iff (ac,b)=1$? Is it true that $(a,b)=(c,b)=1\iff (ac,b)=1$ ? I denote $(a,b):=\gcd(a,b)$.
I tried using Chinese remainder, but not conclusive : For the implication :
$$\mathbb Z/(ab)\cong \mathbb Z/(a)\times \mathbb Z/(b)$$
and $$\mathbb Z/(cb)\cong \mathbb Z/(a)\times \mathbb Z/(c).$$
How will it prove that $\mathbb Z/(acb)\cong \mathbb Z/(ac)\times \mathbb Z/(b)$ ? 
 A: Yes it's true, and you can prove it easily : For the implication, suppose that $(ac,b)\neq 1$. Then, there is $d>1$ s.t. $d\mid ac$ and $d\mid b$. Since $d\mid b$ and $(a,b)=1$, by Gauss lemma, $d\mid c$ what is a contradiction with $(c,b)=1$.
The converse can be proved with the same technic.
A: This is pretty straightforward from Bezout.

$(\implies)$ Suppose $(a,b)=(c,b)=1$. Then there exist non-zero $\alpha,\beta,\gamma,\delta$ such that
$$\alpha a+\beta b=1$$
$$\gamma c+ \delta b=1$$
Multiplying the two equations together yields 
$$(\alpha \gamma) ac+(\alpha \delta a +\beta \delta b +\beta \gamma c)b=1$$
so $(ac,b)=1$.

$(\impliedby)$ Suppose $(ac,b)=1$. Then there exist non-zero $\alpha,\beta$ such that
$$\alpha ac +\beta b=1$$
The fact that 
$$(\alpha a)c+\beta b=1$$
$$(\alpha c)a+\beta b=1$$
yields $(a,b)=(c,b)=1$.
A: Hint $\ (a,b)=1=(c,b)\iff a^{-1},c^{-1}$ exist mod $b$ $\iff (ac)^{-1}$ exists mod $b$ $\iff (ac,b)=1$
A: Another approach is to use the Bezout theorem: $(a,b)=1\iff \exists u,v\in\Bbb{Z},\,\,au+bv=1$. And so we have $au+bv=1$ and $cw+bt=1$. Multiplying those two identities and factoring $b$, one gets
$$acuw+b(aut+vcw+bvt)=1$$
And we have found $r=uw,s=aut+vcw+bvt\in \Bbb{Z}$ such that $acr+bs=1$ and this implies $(ac,b)=1$
