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This is exercise 10 Chapter 1 the book Introduction to Analytic Number Theory by Tom M. Apostol. All alphabets represent integers and by $(w,z)=g$ we mean the greatest common divisor of w and z is g. What I tried:
$m = ax + by$ and $n = cx + dy$. By solving for $x$ and $y$, and considering $ad-bc = \pm 1$ I just got that $(x,y) = (dm-bn, an-cm)$. I can't go further, please help!

Martin Sleziak
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3 Answers3

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The gcd of $x,y$ divides both $m$ and $n$ since they are both linear combinations of $x,y$. Hence it also divides the gcd of $m,n$.

We have $dm-bn=(ad-bc)x=\pm x$ and $-cm+an=(-bc+ad)y=\pm y$, so the gcd of $m,n$ divides both $x$ and $y$ and hence also their gcd.

Since they divide each other the two gcds must be equal.

almagest
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$m$ and $n$ are expressed as linear combinations of $x$ and $y$.

This means $(x,y) \mid m$ and $(x,y) \mid n$, which implies $(x,y) \mid (m,n)$

Treating $m = ax + by$ and $n = cx + dy$ as a system of linear equations, solving gives

$$x = \frac{dm-bn}{ad-bc} \:\: \text{and} \:\: y = \frac{an-cm}{ad-bc}$$

And, since $ad-bc = \pm 1$, then $x = \pm (dm-bn)$ and $y = \pm (an-cm)$

Applying the argument from above, we conclude $(m,n) \mid (x,y)$.

This can only happen when $|(x,y)| = |(m,n)|$,

and since gcd's are positive $(x,y) = (m,n)$

JKnecht
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Following my suggestion in the comments, suppose without loss that $(x,y)=1$ (divide both equations by this if necessary). We have $$ \begin{bmatrix} m \\ n \end{bmatrix} = \begin{bmatrix}a & b \\c & d \end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} $$ By hypothesis the matrix $\begin{bmatrix}a & b \\c & d \end{bmatrix}$ has inverse $\begin{bmatrix}e & f \\g & h \end{bmatrix}$. Hence $$\begin{bmatrix}e & f \\g & h \end{bmatrix} \begin{bmatrix} m \\ n \end{bmatrix} = \begin{bmatrix}x \\ y\end{bmatrix}. $$

In other words, there are integers $e,f,g,h$ such that $x=em+fn$ and $y=gm+hn$. Now, $(m,n)$ divides both $x,y$ and so $(m,n)|(x,y)$. But $(x,y)=1$ so $(m,n)=1$.

vadim123
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