Let $a$ modulo prime $p$ have order 4. Show that
$$ (a + 1)^4 \equiv -4 \pmod p $$
We first expand $(a+1)^4$,
$$ (a+1)^4 = a^4 + 4 a^3 + 6 a^2 + 4 a + 1 $$
We know that $a^4 \equiv 1 \pmod p$. We also know that $a^2 \equiv -1 \pmod p$, because the possibility $a^2 \equiv 1$ is ruled out by $a$ having the order 4 modulo $p$.
We also note that $4a^3 + 4a \equiv 4a(a^2+1) \equiv 4a(-1 + 1) \equiv 0 \pmod p$.
The allows us to reduce the terms as follows
$$ \begin{align} (a+1)^4 & \equiv a^4 + 4 a^3 + 6 a^2 + 4 a + 1 \pmod p \\ \\ & \equiv (1) + 4 a^3 + 6 (-1) + 4 a + 1 \pmod p \\ \\ & \equiv (1) + 6 (-1) + (0) + 1 \pmod p \\ \\ & \equiv -4 \pmod p \end{align}$$
And so
$$ (a + 1)^4 \equiv -4 \pmod p $$