TITLE: Find integers s and t such that $1=7 \times s+11 \times t$. Show that s and t are not unique
QUESTION [7 upvotes]: Find integers s and t such that $1=7 \times s+11 \times t$. Show that s and t are not unique.
I understand why s and t are not unique, I am just unsure how to prove it.
Thank you!

REPLY [8 votes]: If $7s+11t=1$, then $7(s+11k)+11(t-7k)=1$ for all $k$. So there are infinitely many choices.
As for finding one pair $(s,t)$ that work, though there are general procedures, for such small numbers it is easier to go by experimentation. Note that we can take $s=8$ and $t=-5$.
Remark: Suppose that we know that $7s_0+11t_0=1$. We want to find all $(s,t)$ such that $7s+11t=1$. If $s$ and $t$ have this property, then $7s+11t=7s_0+11t_0$.
this can be rewritten as 
$$7(s-s_0)=11(t_0-t).$$
The prime $11$ divides the right-hand side. Since $11$ does not divide $7$, it divides $s-s_0$. Let $s-s_0=11k$. Then from $(7)(11k)=11(t_0-t)$ we get $t=t_0-7k$. So in fact all solutions are of the shape $s=s_0+11k$, $t=t_0-7k$.

REPLY [2 votes]: Hint: To show that it is not unique, all you need to do is find two distinct pairs $(s_1,t_1)$ and $(s_2,t_2)$ such that $1=7s_1+11t_1=7s_2+11t_2$.