Bézout Identity with coefficient divisibility constraints $(a,b)=1$ and $(b,c)=r$. Show that there are two coprime integers $x$ and $y$ satisfying $xa+yb=r$ and $c$ divides $x$.
I got $ua+vb=1$, then $uca+vcb=c$, then $muca+mvcb+nb=mc+nb=r$. But how to prove that $(muc,mvc+n)=1$?
 A: Put $\,d=r,\, e = c\,$ in this enhancement of the Bezout gcd identity.
Theorem $\ \ (c,b) = d\,\Rightarrow\, c\,x + b\,y = d\ $  for some $\,x,y\in\Bbb Z\,$ such that
$\qquad\qquad\quad x\,$ is divisible by any given integer $\,a\neq 0\,$ with $\,\smash{\color{#0a0}{{\large (}a,{\large \frac{b}d}{\large )} = 1}}$
$\qquad\qquad\quad y\,$ is coprime to both $\,x\,$ and  any given integer $\,e\neq 0\,$
Proof $\ \ \color{#0a0}{1}\:\! d = \color{#0a0}{{\large(}a,{\large \frac{b}d}\large{)}}(c,b) = (ac,\,b{\large (}\color{#0a0}{{a,{\large \frac{b}d}}},{\large \frac{c}d}{\large)\:\!\!)}\! = (ac,b).\,$ Let $\, c,e\to ac,ae\,$ in
Lemma $\, (c,b)=d\,\Rightarrow\, c\,x+b\,y=d,\ (ex,y)=1\,$ for some  $\,x,y\in\Bbb Z.\  $ Proof:
$\Rightarrow\ \,c\,x_{\!\circ }+ b\,y_{\circ}  = d,\ $ for $\ x_{\!\circ },y_{\circ }\in\Bbb Z\,$ by Bezout.  $ $ Thus canceling $\,d\,$ yields
$\Rightarrow\ \:\! \bar c\, x_{\!\circ } + \bar b\, y_{\circ}\!\!\: =\, 1,\ $ for $\ \ \bar c = c/d,\ \ \bar b = b/d\qquad\qquad\qquad\qquad\quad\ \ \ [1]$
$\Rightarrow\ \:\! \bar c\,\color{#c00}{x}\ + \bar b\,y\ =\, 1,\ $ for $\ \ x = x_{\!\circ }\!+k\, \bar b,\ \ y = y_{\circ }\! - k\,\bar c,\ \ k\in\Bbb Z\qquad\ \ \ \ \ [2]$
Hence $\ \:\! (\color{#c00}{x },\  y)\, =\,1,\ $ hence $\ {1\!=\!(e\color{#c00}x,y)}\!$ $\iff\! 1 = (\color{c00}e,y) = (e,\,y_{\circ }\! - k\,\bar c)$
true for some $\,k\in\Bbb Z\ $ by  Stieltjes, $ $ by $\:\![1]\Rightarrow (y_{\circ },\bar c)=1$ $\,\Rightarrow\,(e,\,y_{\circ },\,\bar c)\!=\!1$
So $\ \ {c\,x + b\, y\: =\: d\,}\ $ by scaling $\,[2]\,$ by $\,d$.  $\smash{\small\ \ {\bf QED}^{ 2}}$

Remark $\, $ Necessarily $\,\color{#0a0}{(a,b/d)\! =\! 1}\ $ by $\ a\mid x\,$ and $\,(x,b/d)\! =\! (x,\bar b)\! =\! 1\,$ by $[2],\,$ and
$\,(y,x)\! =\! 1\,$ by $[2],\,$ so the Theorem's hypotheses are as general as possible of that sort.
Implicit in the first lines of the proof of the Theorem and Lemma is the following
Lemma' $\ \, c\,x+b\,y = k\,$ has roots $\,x,y\in\Bbb Z\,$ with $\,a\mid x\iff (ac,b)\mid k$
Proof $\ \, $ LHS $\iff c\,a\,\bar x + b\,y = k\,$ has roots $\,\bar x,y\in\Bbb Z\iff (ca,b)\mid k,\,$ by Bezout.
A: I'm not sure offhand about how to finish your proof, but here is an alternate method to consider. Note $\gcd(b,c) = r$ means
$$b = rd, \; c = re, \; \gcd(d,e) = 1 \tag{1}\label{eq1A}$$
Thus, Bézout's identity states there are integers $u$ and $v$ such that
$$u e + v d = 1 \tag{2}\label{eq2A}$$
Also, the general solution is $u_k$ and $v_k$ where, for any integer $k$, you have
$$u_k = u + kd \tag{3}\label{eq3A}$$
$$v_k = v - ke \tag{4}\label{eq4A}$$
Let
$$\gcd(e,r) = f, \; e = fg, \; r = fh, \; \gcd(g,h) = 1 \tag{5}\label{eq5A}$$
$$\gcd(v,h) = m, \; h = n(mq), \; \gcd(m,n) = 1 \tag{6}\label{eq6A}$$
In \eqref{eq6A}, note $q$ contains all prime factors of $m$ in $\frac{h}{m}$ to allow $n$ to be such that $\gcd(m,n) = 1$.
Next, consider the following $2$ modulo equations
$$u_k = u + kd \equiv 0 \pmod a \tag{7}\label{eq7A}$$
$$v_k = v - ke \equiv v + nf \pmod r \tag{8}\label{eq8A}$$
Since $\gcd(a,b) = 1$, \eqref{eq1A} gives $\gcd(a,r) = 1$ and $\gcd(a,d) = 1$. Thus, $d$ has an inverse modulo $a$, so \eqref{eq7A} can be solved by $k \equiv -ud^{-1} \pmod a$. Also, with \eqref{eq8A}, $-ke \equiv nf \pmod r \iff -kg \equiv n \pmod h \iff k \equiv -g^{-1}n \pmod h$. This shows there are solutions for $k$ in \eqref{eq8A} as well. Since $\gcd(a,r) = 1$, as mentioned earlier, then the Chinese remainder theorem shows there exists a solution for $k$, modulo $ar$, of the $2$ congruence equations simultaneously.
Choose one of these values of $k$. From \eqref{eq7A}, for some integer $j$, you have
$$u_k = aj \tag{9}\label{eq9A}$$
Use \eqref{eq9A}, along with the corresponding $v_k$, to replace $u$ and $v$ in \eqref{eq2A}, to get
$$\begin{equation}\begin{aligned}
aj(e) + v_k(d) & = 1 \\
(je)a + v_k(d) & = 1 \\
(jre)a + v_k(rd) & = r \\
(jc)a + (v_k)b & = r
\end{aligned}\end{equation}\tag{10}\label{eq10A}$$
This equation is of the form requested. All that is left is to prove $\gcd(jc,v_k) = 1$. Well, the first line of \eqref{eq10A} shows that $\gcd(j,v_k) = 1$, so just need to show $\gcd(c,v_k) = 1$. Since the first line of \eqref{eq10A} shows $\gcd(e,v_k) = 1$, and $c = re$, then need to show $\gcd(r,v_k) = 1$. As \eqref{eq8A} shows $v_k \equiv v + nf \pmod r$, note that as \eqref{eq2A} shows $\gcd(v,e) = 1$, then \eqref{eq5A} shows $\gcd(v,f) = 1$, then $\gcd(f, v + nf) = 1$.
Thus, since $r = fh$, just need to show $\gcd(v + nf, h) = 1$. Consider any prime factor $p$ of $h$. If $p \mid m$, then $p \mid v$, but $p \not\mid n$, and $p \not\mid f$ (since $p \mid v$ and $\gcd(v,f) = 1$), so $p \not\mid v + nf$. If, instead, $p \mid n$, then $p \mid nf$, but $p \not\mid v$, so $p \not\mid v + nf$. This shows that $\gcd(v + nf, h) = 1$.
This finishes showing you can have
$$xa + yb = r \tag{11}\label{eq11A}$$
by using $x = jc$ and $y = v_k$ from \eqref{eq10A}, and with $\gcd(x,y) = 1$.
