How to calculate all the four solutions to $(p+5)(p-1) \equiv 0 \pmod {16}$? This is a kind of a plain question, but I just can't get something.
For the congruence and a prime number $p$: $(p+5)(p-1) \equiv 0\pmod {16}$.
How come that the in addition to the solutions 
$$\begin{align*}
p &\equiv 11\pmod{16}\\
p &\equiv 1\pmod {16}
\end{align*}$$ 
we also have
$$\begin{align*}
p &\equiv 9\pmod {16}\\
 p &\equiv 3\pmod {16}\ ?
\end{align*}$$
Where do the last two come from? It is always 4 solutions? I can see that they are satisfy  the equation, but how can I calculate them?
Thanks

$$(p+5)(p-1) \equiv 0 \text{ (mod 16)} $$
$$\Leftrightarrow (p+5)(p-1)=16k $$
$$\Leftrightarrow p^{2}+4p+(-5-16k)=0$$
$$\Rightarrow p=-2 \pm\frac{1}{2}\sqrt{36+64k}$$
$$\Rightarrow p=-2 \pm \sqrt{9+16k}$$
$$k=0: p=-2\pm 3 \Rightarrow p \in \{1,-5\} \equiv \{1,11\} \text{ (mod 16)}$$
$$k=1: p=-2 \pm 5 \Rightarrow p \in \{3,-7\} \equiv \{3,9\} \text{ (mod 16)}$$
No other values of $k$ will yield unique integer solutions for $p \text{ (mod 16)}$. To see this, note that in order for $p$ to be an integer, we need $9+16k=q^{2}$ for some $q$. But mod 16, this reduces to $q^{2} \equiv 9 \text{ (mod 16)}$. Now a quick check shows that $q=3$ and $q=5$ are the only solutions. These correspond to $k=0$ and $k=1$, which yield the only 4 possible values: $p \in \{1,3,9,11\}$.
