Show that $a^2 \bmod b = (a \bmod b)^2 \bmod b$ Show that 
$$a^2 \mathrm{\ mod \ } b = (a \mathrm{\ mod \ } b)^2 \mathrm{\ mod\ } b$$
for $ a, b \in \mathbb{Z}^+ $.

this was derived from an Informatics olympiad question.
$$a=kb+n\\a^2=k^2b^2+2kbn+n^2\\a^2=n^2\mod{b}$$
Now since $a=n \mod{b}$ this is the proof
The argument of @kingW3 is sufficient.

By definition of integer division, $a = b\ (a \ div\ b)+(a\ mod\ b)$.

Then $$a^2\ mod\ b=\left(b(a\ div\ b)+(a\ mod\ b)\right)^2\ mod\ b=\left(b^2(a\ div\ b)^2 + 2 b(a\ div\ b)(a\ mod\ b)+(a\ mod\ b)^2\right)\ mod\ b = (a\ mod\ b)^2\ mod\ b.$$