1. ## GCD prof

I need a little help on this one.....Prove that GCD(a,b) = GCD(b, a+b).

2. Originally Posted by Syndicate01
I need a little help on this one.....Prove that GCP(a,b) = GCD(b, a+b).
Suppose that $d\mid a,b$ then $d\mid a+b\implies d\mid a,a+b$ so $(b,a+b)\mid (a,b)$. Work simillarly

3. In general, $\mbox{GCD}(a,b) = \mbox{GCD}(\alpha a+\beta b,\gamma a + \eta b)$ for any matrix $\begin{pmatrix}
{\alpha}&{\beta}\\
{\gamma}&{\eta}
\end{pmatrix} \in \mbox{SL}_2(\mathbb{Z})$
.

4. Originally Posted by Bruno J.
In general, $\mbox{GCD}(a,b) = \mbox{GCD}(\alpha a+\beta b,\gamma a + \eta b)$ for any matrix $\begin{pmatrix}
{\alpha}&{\beta}\\
{\gamma}&{\eta}
\end{pmatrix} \in \mbox{SL}_2(\mathbb{Z})$
.
Heu ...

5. Originally Posted by Syndicate01
Heu ...
$\alpha,\beta,\gamma,\eta\in\mathbb{Z},\text{ and }\alpha\eta-\beta\gamma=1$

6. Originally Posted by Drexel28
$\alpha,\beta,\gamma,\eta\in\mathbb{Z},\text{ and }\alpha\eta-\beta\gamma=1$
$=\pm 1$

7. Originally Posted by Bruno J.
$=\pm 1$
Special linear group - Wikipedia, the free encyclopedia

$\pm 1\implies O_2(\mathbb{Z})$

Replace $\mbox{SL}_2$ by $\mbox{O}_2$ in my post!