Competition problem — Number theory

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Number theory Difficulty 4.6 AIME Prove it Taiwan

Let $n$ be a positive integer greater than 1 and $p$ be a prime such that $n$ divides $p-1$ and $p$ divides $n^3-1$ .
Prove that $4p-3$ is a perfect square.

Solution

Since $n|(p-1)$ , $n < p$ . Since $p|(n^3-1) = (n-1)(n^2+n+1)$ , $p|(n^2+n+1)$ .
Let $p-1 = n\ell$ . Also
$n^2 + n + 1 = n^2 + n + p - n\ell = n(n + 1 - \ell) + p$
so we obtain $p|n(n+1-\ell)$ . Since $n < p$ , $p|(n+1-\ell)$ . Also since $\ell > 0$ , $n+1-\ell \le n$ . Now
$p|(n+1-\ell)$ and $n < p$ , so we get $\ell \ge n+1$ . If $\ell > n+1$ , then $p = 1+n\ell > 1+n+n^2$
which is impossible! since $p|(n^2+n+1)$ . So $\ell = n+1$ and $p = n^2 + n + 1$ . Therefore
$4p-3 = 4(n^2 + n + 1) - 3 = (2n + 1)^2.$

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.