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Problem 871

AMC 12 late, AIME early

Number theory Difficulty 4.7 Prove it Greece — Selection Examination · Greece

If $a$ is an even positive integer and $A = a^n + a^{n-1} + \dots + a + 1$ , $n \in \mathbb{N}^*$ , is a perfect square, prove that $a$ is a multiple of $8$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Since $a$ is an even positive integer, it follows that $A$ is odd. Therefore $A$ will be a perfect square of an odd integer, that is
$A = (2\kappa + 1)^2 = 4\kappa^2 + 4\kappa + 1 = 4\kappa(\kappa + 1) + 1,$
where $\kappa$ is a positive integer. Since one of the two integers $\kappa$ and $\kappa + 1$ is even, we have
$\begin{align*} A &= 4\kappa(\kappa + 1) + 1 = 8\rho + 1, \text{ where } \rho \text{ is a positive integer} \\ \Rightarrow A - 1 &= a^n + a^{n-1} + \dots + a = 8\rho \Rightarrow a(a^{n-1} + \dots + a + 1) = 8\rho \\ \Rightarrow 8|a(a^{n-1} + \dots + a + 1) \Rightarrow 8|a, \text{ since } (8, a^{n-1} + \dots + a + 1) = 1. \end{align*}$

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.