Variation of Fermat's Theorem I'm am having trouble coming up with a proof strategy for the following variation of Fermat's Theorem. If the solution is trivial, please forgive me, this is my first encounter with this theorem. I was able to show/understand (with help) the other variations of this theorem that you'd typically see. $[1]_p=[a^{p-1}]_p$ and $[a]_p=[a^p]_p$ where $[1]_p= 1\pmod p$.
We have: If $a\in \mathbb Z$, and $p$ is prime which does not divide $a$, then $[a^p]_p=[a]_p$.
For all $0<r<p$, there exists a unique $0<s<p$ such that $[sa]_p=[r]_p$. 
We need to show existence and uniqueness such that there exists such an $s$ and that there is no more than one such $s$.
What little I could think of was:
$[sa]_p=[r]_p$
$[s]_p[a]_p=[r]_p$
Is this sufficient to show existence?
From here, I would greatly appreciate some direction. Since we have $[a^p]_p=[a]_p$ and $[sa]_p=[r]_p$, are we trying to show $sa=a$ (thus, $s=1$) and $a^p=r$ in some fashion?

Existence. You want to solve the congruence
$$
sa\equiv r\pmod{p}
$$
that is, finding $s$ and $k$ such that $sa-r=kp$. Now, since $p$ doesn't divide $a$, you know that $a$ and $p$ are coprime, so $1=ax+py$ for some integers $x$ and $y$. Then $r=r1=axr+pyr$, and so

 you can take $s=xr$ and $k=yr$.

Uniqueness. If $s_1$ and $s_2$ are solutions, then $(s_2-s_1)a\equiv 0\pmod{p}$. But $p$ doesn't divide $a$, so it must divide $s_2-s_1$.
