Olympiad Corner 1

Though it seems to be difficult but it is very easy.

Let $2^n + 12^n + 2011^n = l^2$ .

Case 1: If $n\geq 2$ .

Since we assumed that $n \geq 2$ , we have $2^{n} \equiv 0 \mod 4$ , $12^n \equiv 0 \mod 4$ and $2011^n \equiv (-1)^n \mod 4$ . Adding all these we get $l^2 \equiv 2^n + 12^n + 2011^n \equiv (-1)^n \mod 4$ . But a perfect square leaves either $0$ or $1$ as remainder when divided with $4$ . So, it gives that $n$ is even and cannot be odd.

As we got that $n$ is even, take $n=2k$ .

Now, $2^{2k} \equiv 1 \mod 3$ , $12^{2k} \equiv 0 \mod 3$ and $2011^{2k} \equiv 1 \mod 3$ . Adding these we get $l^2 \equiv 2^n + 12^n + 2011^n \equiv 2^{2k} + 12^{2k} + 2011^{2k} \equiv 2 \mod 3$ . But a perfect square leaves only $0$ or $1$ as remainder when divided with $3$ . But this expression gives $2$ as remainder when divided with $3$ . So there exists no such $n$ which satisfies the given condition in this case.

Case 2: If $n<2$ i.e. $n=1$ .

It is easy to check this case. Taking $n=1$ , we get $2025$ which is square of $45$ i.e. it is a perfect square. Thus only one solution exists i.e. $\boxed{n=1}$