# 2011 USAJMO Problems/Problem 1

Let . 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 that satisfies is . Assume that . Then consider the equation . From modulo 2, we easily that x is odd. Let , where a is an integer. . 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 , , so similarly, the entire LHS is an integer, and so are and . 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 . . Thus, n is even. However, I have already shown that must be odd. This is a contradiction. Therefore, 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