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
By direct calculation, you can find that $(\pm3)^2\equiv (\pm11)^2\equiv9\mod 16$. The result follows directly.
However, if you are asking for more general situations, you might be interested in quadratic residues.
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
