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


<p/>



<br clear="all" /><table border="0" width="100%"><tr><td>
<table align="center"><tr><td nowrap="nowrap" align="center">
1<sup>5</sup><font face="symbol">-</font
>1,&#x00A0;2<sup>5</sup><font face="symbol">-</font
>2,&#x00A0;3<sup>5</sup><font face="symbol">-</font
>3,<font face="symbol">¼</font
>&#x00A0;<i>n</i><sup>5</sup><font face="symbol">-</font
><i>n</i>.</td></tr></table>
</td></tr></table>




<p/>
The solution depends 
only on a little algebra and some clear mathematical thinking.


<p/>


Pierre Geoghegan, age 17, Tarbert Comprehensive, Ireland, Prateek Mehrotra,
 age 14, Riccarton High School, Christchurch, New Zealand and Vassil Vassilev, age 16, from Lawnswood Sixth Form, Leeds started by 
taking small values of <i>n</i>, usually a good way to begin. This solution comes 
from Arun Iyer, age 17, S.I.A High School and Junior College, India, and 
they all found the answer which is 30.
<p/>


Given the sequence 1<sup>5</sup><font face="symbol">-</font
>1,&#x00A0;2<sup>5</sup><font face="symbol">-</font
>2,&#x00A0;3<sup>5</sup><font face="symbol">-</font
>3,<font face="symbol">¼</font
>&#x00A0;<i>n</i><sup>5</sup><font face="symbol">-</font
><i>n</i> we see that
<p/>





<br clear="all" /><table border="0" width="100%"><tr><td>
<table align="center"><tr><td nowrap="nowrap" align="center">
<i>n</i><sup>5</sup><font face="symbol">-</font
> <i>n</i> = <i>n</i>(<i>n</i><sup>4</sup> <font face="symbol">-</font
> 1) = <i>n</i>(<i>n</i> <font face="symbol">-</font
> 1)(<i>n</i> + 1)(<i>n</i><sup>2</sup> + 1)</td></tr></table>
</td></tr></table>




<p/>


and it is quite easy to see
that <i>n</i>(<i>n</i><font face="symbol">-</font
>1)(<i>n</i>+1)(<i>n</i><sup>2</sup>+1) is divisible by 2, 3 and 5 for all values of
<i>n</i>. As <i>n</i>, (<i>n</i><font face="symbol">-</font
>1) and (<i>n</i>+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 <i>n</i> is either of the form 5<i>k</i>+2 or 5<i>k</i>+3 for some
integer <i>k</i> and in both of these cases we can check that <i>n</i><sup>2</sup> + 1 is 
divisible by 5. Since 2, 3 and 5 are coprime therefore <i>n</i><sup>5</sup> <font face="symbol">-</font
> <i>n</i> is
divisible by 2*3*5 i.e by 30.


<p/>
Since the second term of the sequence is 2<sup>5</sup><font face="symbol">-</font
>2 = 30 therefore
the divisor cannot be greater than 30. Therefore 30 is the largest number that d
ivides each member of the sequence.