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

The assertion $(p+5)(p-1) \equiv 0 \pmod{16}$ is equivalent to $16 \mid (p+5)(p-1)$. Then you consider cases: $2^4 \mid (p+5)$, $2^3 \mid (p+5)$ and $2 \mid p-1$, $2^2 \mid p+5$ and $2^2 \mid p-1$, etc. 
