Prove $a = 2 \cdot 10^{2016} + 16$ cannot be a perfect square 
Prove the number $a = 2 \cdot 10^{2016} + 16$ cannot be a perfect square.

I tried studying $a \mod b$ for different integers $b$, without succes.

Hint $\ $ To prove $\,2\,(10^{\large 2N}\!+8)\,$ nonsquare it suffices to find a prime $p$ where it is nonsquare mod $p,\,$ e.g. if $\,2\,$ is nonsquare and $\,10^{\large 2N}\!\!+8$ is square (e.g. if $\,10^{\large 2N}\!\equiv 1),\,$ e.g. $\,p = 11\,$ (Arthur's answer). 
