If $p, q, r, s$ are integers such that $ps - qr = \pm 1$, and $a, b, a', b'$ are integers such that
$$ a' = pa + qb $$
$$ b' = ra + sb $$
Prove that $\text{gcd}(a', b') = \text{gcd}(a, b)$.
I understand that posting questions without showing your work is a bad thing, however, I don't know how to even start this proof, as I don't see the connection between the given conditions and what I have to prove. Any help is appreciated. Thanks.
This is a cute problem. It's standard to show $\text{gcd}(a,b)\ \big|\ \text{gcd}(a',b')$. For the other direction, it is sufficient to show there exist integers $x,y,z$ and $w$ such that $a = xa'+yb'$ and $b = za'+wb'$.
Here is one way to derive $x,y,z$ and $w$. The relationship between $(a,b)$ and $(a',b')$ can be written as follow: $$\begin{bmatrix}a'\\ b'\end{bmatrix} = \begin{bmatrix}p&q\\ r&s\end{bmatrix}\begin{bmatrix}a\\ b\end{bmatrix}$$ Therefore, $$\begin{bmatrix}a\\ b\end{bmatrix} = \begin{bmatrix}p&q\\ r&s\end{bmatrix}^{-1}\begin{bmatrix}a'\\ b'\end{bmatrix} = \frac{1}{ps-qr}\begin{bmatrix}s&-q\\ -r&p\end{bmatrix}\begin{bmatrix}a'\\ b'\end{bmatrix} = \begin{bmatrix}x&y\\ z&w\end{bmatrix}\begin{bmatrix}a'\\ b'\end{bmatrix}$$ Because $ps-qr = 1$ or $-1$, $x,y,z$ and $w$ are integers.
