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. Unknown error_msg). 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. Unknown error_msg) must be an integer. Let $\dfrac {1}{4} (1 - 2011^n}) = k$ (Error compiling LaTeX. Unknown error_msg). 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

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2011 USAJMO Problems/Problem 1

Solution