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Given $x$ and $y$, let $m=ax+by$ and $n=cx+dy$, where $ad-bc=\pm1$. Prove that $(m,n)=(x,y)$.

First attempt Given that $(x,y)\mid m$ and $(x,y)\mid n$ so $(x,y)\mid (m,n)$. Furthermore, from the initial relations one can find that $$ x=\frac{dm-bn}{ad-bc}=\pm(dm-bn),\quad y=\frac{an-cm}{ad-bc}=\pm(an-cm) $$ but so $(m,n)\mid(x,y)$ by the same reasoning. Being the gcd always positive one may conclude that $(man)=(x,y)$.

Second Try If, $g=(m,n)=(ax+by,cx+dy)$, then there are $\alpha,\beta$ such that $$ (ax+by)\alpha+(cx+dy)\beta=g $$ and rewriting one find that $$ x(a\alpha+c\beta)+y(b\alpha+d\beta)=(x,y)=g. $$

I feel the first is better.

yngabl
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    Your first try is correct. In the second one I cannot see why $x(a\alpha+c\beta)+y(b\alpha+d\beta)=(x,y).$ See also ${\rm gcd}(x,y)$ dividing linear combination of $x$, $y$ – Anne Bauval Jul 25 '23 at 15:38
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    In the Second Try, prepend a $\min_{\alpha,\beta\in\mathbb{Z}}$ on the left of the first equation. Then, since $ad-bc=\pm1$, this is the $a\alpha+c\beta$ and $b\alpha+d\beta$ also range over all integers. So, the left of the second equation is really $\min_{\alpha',\beta'\in\mathbb{Z}}(x\alpha'+y\beta')$, which is then $(x,y)$. – NDB Jul 25 '23 at 15:49

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