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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

# 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 $4p-3=(2n+1)^2$.

# Connected Program at Cheenta

#### Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

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