Prove that for any given integers $b > a \geq 1$ there exists an integer solution $u$, $w$ to $au - bw = \text{gcd}(a,b)$ with $0\leq u\leq b-1$ and $0\leq w \leq a-1$.
This is supposedly a simple exercice in applying the Euclidean division. I tried a lot of things but my technique is mainly starting with any solution $m,n\in\mathbb{Z}$ such that $am-bn=(a,b)$ and then setting $m=qb+u$ for some $q,u\in\mathbb{Z}$ with the constraint that $0\leq u\leq b-1$. Then, by substitution, I get
\begin{equation*} a(qb+u) - bn = au - b(n - qa) = (a,b) \end{equation*}
I don't know what is the best way to proceed here. Do I need to show that $0\leq n-qa\leq a-1$? It seems to me that there must be some quick and elegant way to do this, but I've been playing with inequalities for hours and nothing seems to work.
Am I on the right track or did I miss something important? I'm trying to see a pattern here.