# Common Divisor

Find the largest integer which divides every member of the
following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.

## Problem

Find the largest integer which divides every member of the following sequence:

$$ 1^5-1,\ 2^5-2,\ 3^5-3,\cdots\ n^5-n.$$

## Getting Started

Try more terms of the sequence. Think about factors.

## Student Solutions

The problem was to find the largest integer which divides every member of the following sequence:

\[1^5-1,\ 2^5-2,\ 3^5-3,\cdots\ n^5-n.\]

The solution depends only on a little algebra and some clear mathematical thinking.

Pierre, Tarbert Comprehensive, Ireland, Prateek, Riccarton High School, Christchurch, New Zealand and Vassil from Lawnswood Sixth Form, Leeds started by taking small values of $n$, usually a good way to begin. This solution comes from Arun Iyer, S.I.A High School and Junior College, India. They all found the answer which is $30$.

Given the sequence $1^5-1,\ 2^5-2,\ 3^5-3,\cdots \ n^5-n$ we see that

\[n^5 - n = n(n^4 - 1) = n(n - 1)(n + 1)(n^2 + 1)\]

and it is quite easy to see that $n(n-1)(n+1)(n^2+1)$ is divisible by $2$, $3$ and $5$ for all values of $n$. As $n$, $(n-1)$ and $(n+1)$ are three consecutive integers their product must be divisible by $2$ and by $3$. If none of these numbers is divisible by $5$ then $n$ is either of the form $5k+2$ or $5k+3$ for some integer $k$ and in both of these cases we can check that $n^2 + 1$ is divisible by $5$. Since $2$, $3$ and $5$ are coprime therefore $n^5 - n$ is divisible by $2 \times 3 \times 5$ i.e by $30$.

Since the second term of the sequence is $2^5-2 = 30$ therefore the divisor cannot be greater than $30$. Therefore $30$ is the largest number that d ivides each member of the sequence.

## Teachers' Resources

Why do this problem?

It gives practice in factorisation and an opportunity for learners to make and prove conjectures.

Possible approach

This problem makes a good lesson starter.

Key questions

What are the factors of $n^5 - n$?