Proof that $\gcd(ax+by,cx+dy)=\gcd(x,y)$ if $ad-bc= \pm 1$

I'm having problems with an exercise from Apostol's Introduction to Analytic Number Theory.

Given $x$ and $y$, let $m=ax+by$, $n=cx+dy$, where $ad-bc= \pm 1$. Prove that $(m,n)=(x,y)$.

I've tried to give a proof, but I suspect it's wrong (or at least not very good). I would be very thankful for any hints/help/advice!

My proof:

We observe that since $ad-bc= \pm 1$, $ad=bc \pm 1$, and $(ad,bc)=1$. Now, $(a,b) \mid m$ and $(c,d) \mid n$ but

$$(a,b) = ((d,1)a,(c,1)b)=(ad,a,bc,b)=(ad,bc,a,b)=((ad,bc),a,b)=(1,a,b)=1.$$

Similarly, we determine $(c,d)=1$. So, $1=(a,b) \mid m$ and $1=(c,d) > \mid n$. But $(x,y)$ also divide $m$ and $n$. Since $(x,y) \geq (a,b)=(c,d)=1$, this implies that $(x,y)=m,n$. Hence $(m,n)=((x,y),(x,y))=(x,y)$.

Here is a proof. Call $z=(x,y)$ and $p=(m,n)$. The expressions of $m$ and $n$ as integer linear combinations of $x$ and $y$ show that $z$ divides $m$ and $n$ hence $z$ divides $p$ by definition of the gcd. On the other hand, $\pm x=dm-bn$ and $\pm y=cm-an$ hence the same argument used "backwards" shows that $p$ divides $\pm x$ and $\pm y$, which implies that $p$ divides $z$, end of proof.

Hint: Any common factor of $x$ and $y$ clearly divides both $m$ and $n$, because if $f\mid x$ and $f\mid y$, then also $f\mid ax+by$ et cetera.

The task at hand is to prove the reverse fact: any common factor of $m$ and $n$ also divides $x$ and $y$. One way of seeing this follows from the matrix equation $$ \left(\begin{array}{c}m\\n\end{array}\right)= \left(\begin{array}{cc}a&b\\c&d\end{array}\right)\left(\begin{array}{c}x\\y\end{array}\right). $$

Does anything about the given condition on $ad-bc$ help you with the inverse of the above $2\times2$ matrix?

Hint $ $ (excerpted from my answer to a similar question a few months ago) $ $ Generally, inverting a linear map by Cramer's Rule, i.e. $\rm\color{#0a0}{scaling}$ by the $\rm\color{#c00}{adjugate}$ yields

$$\begin{align} \rm\color{#c00}{ \begin{bmatrix}\rm d\!\! &\!\!\! \rm -b \\ \!\rm -c & \rm a \end{bmatrix}} \color{#0a0}\times&\, \left\{\, \begin{bmatrix}\rm a & \rm b \\ \rm c & \rm d \end{bmatrix} \begin{bmatrix} \rm x \\ \rm y \end{bmatrix} = \begin{bmatrix}\rm X \\ \rm Y\end{bmatrix}\, \right\}\\[.2em] \Longrightarrow &\,\qquad\qquad \begin{array}\ \rm\Delta\ x\ =\, \ \ \rm \color{#c00}d\ X \color{#c00}{- b}\ Y \\ \rm\Delta\ y\ = \rm \color{#c00}{-c}\ X + \color{#c00}a\ Y \end{array}^{\phantom{|}} ,\ \ \ \ \rm \Delta\ :=\ \color{#c00}{ad-bc} \end{align}\qquad$$

Hence $\rm\ n\ |\ X,Y\ \Rightarrow\ n\ |\ \Delta\:x,\Delta\:y\ \Rightarrow\ n\ |\ gcd(\Delta\:x,\Delta\:y)= \Delta\ gcd(x,y)\,$ by here & here.

Hint $ $ Homogeneous reduce to the case $\:(x,y) = 1\:$ by cancelling any gcd. Note Bezout's GCD identity is equivalent to: $\ (\color{#c00}{x,y}) = 1\iff \color{#c00}{\begin{smallmatrix}x\\ y\end{smallmatrix}}\,$ is a column in a matrix of determinant $\,1,\,$ thus

$\ \ \ \ \ \ (\color{#c00}{x,y}) = 1 \ \Rightarrow\ 1 = \begin{vmatrix} a & b \\ c & d\end{vmatrix}\ \begin{vmatrix} \color{#c00}x & u \\ \color{#c00}y & v \end{vmatrix} = \begin{vmatrix} ax+by & s \\ cx+dy & t \end{vmatrix}\,$ $ \Rightarrow\ (ax+by,\ cx+dy) = 1$

The converse follows the same way using the inverse transformation (or by easy arithmetic).

First note that $(x,y)$ divides both $m$ and $n$, hence $(x,y)|(a,b)$. So it suffices to show $(a,b)|(x,y)$

Consider the matrix $T=\left(\begin{array}{cc}a & b \\ c & d\end{array}\right)$. By assumption $\det T=ad-bc=\pm 1$. Hence its inverse satisfies

$T^{-1}=\pm\left(\begin{array}{cc}d & -b \\ -c & a\end{array}\right)$.

Now by definition of $m$ and $n$ we have

$T\left(\begin{array}{c}x\\ y \end{array}\right)=\left(\begin{array}{c}m\\ n \end{array}\right)$.

Hence

$\left(\begin{array}{c}x\\ y \end{array}\right)=T^{-1}\left(\begin{array}{c}m\\ n \end{array}\right)=\left(\begin{array}{c}dm-bn\\ -cm+an \end{array}\right),$

showing that $m$ and $n$ both divide $x$ and $y$.