Given $(a,m)=(b,m)=1$, show : $(ab,m)\mid ab,mb \implies (ab,m)\mid b$

Question

Given $\exists a,m,b \in \mathbb{Z}, (a,m)=(b,m)=1$, show : $(ab,m)\mid ab,mb \implies (ab,m)\mid b$

Attempted proof:
$\exists a,m,b,x,y,x',y'\in \mathbb{Z}$,
(a): $ (a,m)=1 \implies ax +my=1 $; also $(a,m)=1 \implies (ab,mb)=b$
(b): $(b,m)=1 \implies bx' +my'=1$.
Multiplying, (a), (b) get
(c): $abxx' + amxy'+bmx'y+m^2yy'=1$

Am unable to derive from $(ab,mb)=b$, that $(ab,m) \mid ab,mb$ which is a pre-requisite to proof.

Apart from this, have failed to capture the essence of the derivation/proof desired, it would be better if there were an intuitive way to look into it.

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Edit Based on comment by @mathlove, have tried to view (c) as :
(c) $abxx' + m(ax'y'+bx'y+myy') = 1$
=> So, $(ab, m)=1$, with coefficient of $ab= xx'$, & coefficient of $m=ax'y'+ bx'y'+myy'$.

Need prove that $(ab,m) \mid ab,mb$; & it is trivial to prove for $(ab,m) \mid ab$, & similarly for $(ab,m) \mid mb$.

Next, need prove $(ab,m) \mid b$, which for co-prime $ab,m$ is always true. In fact, it is always true for any integer in place of $b$, i.e. $\forall c \in \mathbb{Z}, (ab,m) \mid c$.

However, I have a thought regarding there being not the trivial condition $(a,m)=(b,m)=1$. In that situation, I hope NOTHING can be said about such relationships to occur. May be holding this trivial condition true, forms the needed base to view these relations intuitively.

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Request detail on situation that not considers the stated trivial conditions, i.e. $(a,m)=(b,m)=1$, and then what relationships are possibly derivable, and why or why not.

Answer 1

Since (c) can be written as $$abxx'+m(axy'+bx'y+myy')=1$$ we see that $(a,m)=1$ and $(b,m)=1$ imply $(ab,m)=1$.

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If $(a,m)=1$ and $(b,m)=g\gt 1$, then $(ab,m)\mid b$.

Proof : There exist integers $x,y,x',y'$ such that $ax+my=1$ and $bx'+my'=g$, so$$abxx'+m(axy'+bx'y+myy')=g\implies (ab,m)\mid g\implies (ab,m)\mid b$$

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If $(a,m)=g\gt 1$ and $(b,m)=1$, then $(ab,m)\not\mid b$.

Proof : There exist integers $x,y,x',y'$ such that $ax+my=g$ and $bx'+my'=1$, so$$abxx'+m(axy'+bx'y+myy')=g\implies (ab,m)\mid g\implies (ab,m)=g$$Supposing that $(ab,m)\mid b$ implies $g\mid b$ and $(b,m)\ge g$ which is impossible. So, $(ab,m)\not\mid b$.

Answer 2

I think this will help you,

let $a,b,d \in \mathbb{Z}$, where $d$ and $a$ are nonzero.

if $d\mid ab$ and $p=gcd(a,d)$ then $\frac{d}{p}\mid b$.

which is easy to prove.