# 2011 USAJMO Problems/Problem 1

Find, with proof, all positive integers $n$ for which $2^n + 12^n + 2011^n$ is a perfect square.

## Solution

Let $2^n + 12^n + 2011^n = x^2$ $(-1)^n + 1 \equiv x^2 \pmod {3}$. Since all perfect squares are congruent to 0 or 1 modulo 3, this means that n must be odd. Proof by Contradiction: I will show that the only value of $n$ that satisfies is $n = 1$. Assume that $n \ge 2$. Then consider the equation $2^n + 12^n = x^2 - 2011^n$. From modulo 2, we easily that x is odd. Let $x = 2a + 1$, where a is an integer. $2^n + 12^n = 4a^2 + 4a + 1 - 2011^n$. Dividing by 4, $2^{n-2} + 3^n \cdot 4^{n-1} = a^2 + a + \dfrac {1}{4} (1 - 2011^n})$ (Error compiling LaTeX. ! Extra }, or forgotten $.). Since $n \ge 2$, $n-2 \ge 0$, so $2^{n-2}$ similarly, the entire LHS is an integer, and so are $a^2$ and $a$. Thus,$\dfrac {1}{4} (1 - 2011^n})$(Error compiling LaTeX. ! Extra }, or forgotten$.) must be an integer. Let $\dfrac {1}{4} (1 - 2011^n}) = k$ (Error compiling LaTeX. ! Extra }, or forgotten \$.). Then we have $1- 2011^n = 4k$. $1- 2011^n \equiv 0 \pmod {4}$ $(-1)^n \equiv 1 \pmod {4}$. Thus, n is even. However, I have already shown that $n$ must be odd. This is a contradiction. Therefore, $n$ is not greater than or equal to 2, and must hence be less than 2. The only positive integer less than 2 is 1. -hrithikguy

## Solution 2

If $n = 1$, then $2^n + 12^n + 2011^n = 2025 = 45^2$, a perfect square.

If $n > 1$ is odd, then $2^n + 12^n + 2011^n \equiv 0 + 0 + (-1)^n \equiv 3 \pmod{4}$.

Since all perfect squares are congruent to $0,1 \pmod{4}$, we have that $2^n+12^n+2011^n$ is not a perfect square for odd $n > 1$.

If $n = 2k$ is even, then $(2011^k)^2 < 2^{2k}+12^{2k}+2011^{2k}$ $= 4^k + 144^k + 2011^{2k} <$ $2011^k + 2011^k + 2011^{2k} < (2011^k+1)^2$.

Since $(2011^k)^2 < 2^n+12^n+2011^n < (2011^k+1)^2$, we have that $2^n+12^n+2011^n$ is not a perfect square for even $n$.

Thus, $n = 1$ is the only positive integer for which $2^n + 12^n + 2011^n$ is a perfect square.

## Solution 3

Looking at residues $mod 3$, we see that $n$ must be odd, since even values of $n$ leads to $2^n + 12^n + 2011^n = 2 \pmod{3}$. Also as shown in solution 2, for $n>1$, $n$ must be even. Hence, for $n>1$, $n$ can neither be odd nor even. The only possible solution is then $n=1$, which indeed works.