Competition problem — Number theory

Library / /122 of 520

Number theory Difficulty 5.7 AIME, harder Prove it

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

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 $\delta_{p}(a+1)=6$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.