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## 'Common Divisor' printed from http://nrich.maths.org/

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.