Track / Stage 5 / 334 of 400 #934 of 1964

Problem 934

AIME late

Number theory Difficulty 5.7 Prove it

Let $p$ be a prime, $a \in \mathbf{N}^{*}$ . Prove: If $\delta_{p}(a)=3$ , then
$\delta_{p}(a+1)=6$

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

1. From $\delta_{p}(a)=3$ , we know $a \neq \pm 1(\bmod p)$ , and $a^{2}+a+1 \equiv 0(\bmod p)$ . Therefore, $1+a \not \equiv 1(\bmod p),(1+a)^{2}=1+2 a+a^{2} \equiv a \not \equiv 1(\bmod p),(1+a)^{3} \equiv$ $(1+a) a \equiv-1(\bmod p)$ , so $\delta_{p}(a+1)=6$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.