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
 A: 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 $a \mid (v_1-v_2)$. 
A: 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.
A: Hint $\ \ \color{#0a0}a\mid \color{#c00}{b(v_2-v_1\!)}\,\Rightarrow\, \color{#0a0}a\mid \overbrace{(\color{#0a0}a\,u_1+v_1\color{#c00}b)}^{\large =\ 1}(\color{#c00}{v_2-v_1}\!).\ \, $  Alternatively use Euclid's Lemma.
Remark $\ $ This shows that $\ a\mid b\color{}c\,\Rightarrow\,a\mid\gcd(a,b)\color{}c.\,$ Yours is special case $\,\gcd(a,b)=1$.
This will be clarified arithmetically when your learn modular arithmetic, where it can be rewritten as follows. If $\,\gcd(a,b)=1\,$ then $\,b^{-1}$ exists $\,$ mod $a\,$ so mutiplying $\,bc\equiv 0\,$ by $\,b^{-1}$ yields $\,c\equiv 0,\,$ i.e. $\,b\,$ is invertible so cancelllable. Above the Bezout identity yields the inverse, i.e. by Bezout $\,\gcd(a,b)=1$ $\,\Rightarrow\,$ $\, au_1\!+ v_1 b = 1,\,$ so, $\,$ mod $\,a\!:\ v_1b\equiv 1$ $\,\Rightarrow\,$  $b^{-1}\equiv v_1$.
