Integers divide several solutions to Greatest Common Divisor equation I'm not sure about the topic's correctness but my problem is following:
Suppose $u_1,v_1$ and $u_2,v_2$ are two different solutions for $au_i + bv_i = 1$, then $a \mid v_2-v_1$ and $b\mid u_1-u_2$.
Well, I have tried to prove this without success, but here are some of my thoughts so far. I want to show that $(v_2-v_1)=ak$ for some $k\in \mathbb{Z}$, and I also know that $au_1+bv_1 = au_2+bv_2 \implies a(u_1-u_2) = b(v_2-v_1)$. In this last equality I know that $gcd(a,b)=1$ from the initial assumption, but what can I say about $gcd((v_2-v_1),(u_1-u_2))$ ? Does that help me in any way?
Best regards

If $u_1,v_1$ and $u_2,v_2$ are solutions to $au_i+bv_i=1$, then $$0=au_1+bv_1-au_2-bv_2=a(u_1-u_2)+b(v_1-v_2).$$
Thus $a \mid b(v_1-v_2)$. Now use $(a,b)=1$ to conclude $a \mid (v_1-v_2)$. 
