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# 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

Number Theory
Easy
##### Suggested Book
An Excursion in Mathematics

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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