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