If $a^2 - 4a \equiv - 3 \pmod{p}$ then $a$ is congruent 1 or 3 modulo $p$ 
Assume $p$ is a prime and $a$ is a number such that 
  \begin{equation}
a^2 - 4a \equiv - 3 \pmod{p}
\end{equation}
  Show that then we must either have $a \equiv 3 \pmod{p}$ or $a \equiv 1 \pmod{p}$. 

If we assume $a \equiv 3 \pmod{p}$ we get 
\begin{align*}
a^2 - 4a = a(a-4) \equiv 3(3-4) \equiv - 3 \pmod{p}
\end{align*}
Ans if we assume  $a \equiv 1 \pmod{p}$ we get
\begin{align*}
a^2 - 4a = a(a-4) \equiv 1(1-4) \equiv - 3 \pmod{p}
\end{align*}
So $a^2 - 4a \equiv - 3 \pmod{p}$ seems to hold for both cases $a$ congruent to 1 or 3 modulo $p$. But this doesn't really prove anything. How can I prove that these two cases are the only ones which fulfills $a^2 - 4a \equiv - 3 \pmod{p}$?

Hint: 
$
a^2 - 4a \equiv - 3 \bmod{p} \iff a^2 - 4a + 3 \equiv 0 \bmod{p}
$.
Now $a^2 - 4a + 3=(a-1)(a-3)$.
