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.
Note that we have
$$
2\cdot 10^{2016} + 16 \equiv 2\cdot(-1)^{2016} + 5 = 7\pmod {11}
$$
but the only squares modulo $11$ are $0, 1, 3, 4, 5, 9$.
We have

$$\sqrt2\cdot10^{1008}\le\sqrt{2\cdot10^{2016}+16}=\sqrt2\cdot10^{1008}\sqrt{1+8\cdot10^{-2016}}&lt;\sqrt2\cdot10^{1008}(1+4\cdot10^{-2016})$$ or

$$0\le\sqrt{2\cdot10^{2016}+16}-\sqrt2\cdot10^{1008}&lt;\sqrt2\cdot4\cdot10^{-1008},$$
which is equivalent to

$$\{\sqrt2\cdot10^{1008}\}&lt;\sqrt2\cdot4\cdot10^{-1008}.$$
This would imply that the decimals of $\sqrt2$ are all zeroes from the $1008^{th}$ to the $2015^{th}$.

So $2\cdot10^{2016}+16$ is probably not a square. :-)