if $\displaystyle (a,m)=1$, then $\displaystyle ax\equiv 1 \ (\text{mod} \ m)$ has a unique sol.

By Bezout, there exist $\displaystyle x,y\in\mathbb{Z}$ such that $\displaystyle 1=ax+my$

Now, we can write that as:

$\displaystyle my=1-ax\Rightarrow m(-y)=ax-1\Rightarrow ax\equiv 1 \ (\text{mod} \ m)$

Correct?