$ad-bc=1$, $w=au+bv$ and $z=cu+dv$. Prove $\gcd(u,v) = \gcd (w,z)$.

Question

As indicated indicated in the title,I am solving the following problem: If $ad-bc=1$, $w=au+bv$ and $z=cu+dv$, prove $\gcd(u,v) = \gcd (w,z)$

So from $ad-bc=1$ I was able to found out $\gcd(a+b,c+d) = 1$.

From here I'm sort of lost as what to do with $w = au + bv$ and $z = cu + dv$.

Thanks for your help.

$$\gcd(w,z) = \gcd(au + bv,cu + dv)$$

Knowing that $\gcd(a+b,c+d) = 1$ how can I move forward from the step above.

Answer 1

This is a simple case of looking at the matrix

$$M = \begin{pmatrix}a & b \\ c & d\end{pmatrix}\in SL_2(\Bbb Z)$$

Then $M\cdot \left(\begin{smallmatrix} u \\ v \end{smallmatrix}\right) =\left(\begin{smallmatrix} w \\ z \end{smallmatrix}\right)$. But then since $M^{-1}\cdot \left(\begin{smallmatrix} w \\ z \end{smallmatrix}\right)= \left(\begin{smallmatrix} u \\ v \end{smallmatrix}\right)$ we see each set of variables are linear combinations of the others, hence each gcd divides the other, hence they are equal.