Sum of N squared is Perfect Square

Let $s^2 = 2^n + 12^n + 2011^n$

In mod $3$ , this becomes

$s^2 \equiv 2^n+1 \equiv (-1)^n+1 \; (mod \; 3)$

Perfect Squares in mod $3$ are congruent to $0$ or $1$

So $n$ must be odd

Now we take mod $4$

$s^2 \equiv 2^n+3^n \equiv 2^n+(-1)^n \; (mod \; 4)$

For odd $n\geq 2, \; 4 \: | \: 2^n$ so $s^2 \equiv (-1)^n\equiv -1\equiv 3 \; (mod \; 4)$

But Squares of mod $4$ are $0$ or $1$

Thus, the only possible solution is $\boxed{n=1}$

Substituting gives $2^1+12^1+2011^1=2025={45}^2$