Understand the problem

Let $ n$ and $ p$ be positive integers greater than $ 1$, with $ p$ being a prime. Show that if $ n$ divides $ p-1$ and $ p$ divides $ n^3-1$, then $ 4p-3$ is a perfect square.

Source of the problem

Argentina TST 2005

Topic
Number Theory
Difficulty Level
Easy
Suggested Book
An Excursion in Mathematics

Start with hints

Do you really need a hint? Try it first!

Note that, p divides either n-1 or n^2+n+1. Using inequalities, show that p has to be greater than n-1. Hence p|n^2+n+1.
Write n^2+n+1=kp for some integer k. Using the fact that n|p-1, show that the integer k has residue 1 modulo n.
Note that, if p\neq n^2+n+1, then k has to be greater than or equal to n+1. Show that this leads to a contradiction.
As p has residue 1 modulo n, we have $p\ge n+1$. This means that n^2+n+1=kp\ge (n+1)(n+1)=n^2+2n+1. This is clearly a contradiction. Hence p=n^2+n+1. Therefore 4p-3=(2n+1)^2.

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