How to reduce a polynomial congruences Consider the Legendre Symbol 
(2|p)
which  give the congruences
$2^{\frac{p-1}{2}} = (-1)^{\frac{p^2-1}{8}} mod p$.
Now ${\frac{p^2-1}{8}}$ is odd if is equal to 2k+1 with k integer
that gives $p^2 = 16 k + 9$ and brings to the polynomial congruences
$p^2 \equiv 9 (mod \,\,\,16)$.
Now the solution gives the congruences 
$p \equiv \pm 3 (mod \,\,\,8)$ so a reduction is possible. Do you know why?
Thanks

If $2^{m+2}|(p-a)(p+a),m\ge1$ and $a$ odd
As $p+a,p-a$ have the same parity, both must be even
$\implies2^m|\dfrac{p-a}2\cdot\dfrac{p+a}2$ 
As the two multipliers have opposite parities
$2^m$ will divide exactly one of them
