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

Hint:
$$au_i+bv_i=1$$
can be rewritten as
$$au_i = 1-bv_i$$
so you have the two equations 
$$au_1 = 1-bv_1$$
$$au_2 = 1-bv_2$$
from which follows that
$$a(u_1-u_2)=b(v_2-v_1)$$
You already know that $\gcd(a,b)=1$, you can use the equation above to deduce the divisibility you need.
