The problem was to find the largest integer which divides every member of the
following sequence:
15-1, 25-2, 35-3,¼ n5-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 15-1, 25-2, 35-3,¼ n5-n we see that
n5 - n = n(n4 - 1) = n(n - 1)(n + 1)(n2 + 1)
and it is quite easy to see
that n(n-1)(n+1)(n2+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 n2 + 1 is
divisible by 5. Since 2, 3 and 5 are coprime therefore n5 - n is
divisible by 2*3*5 i.e by 30.
Since the second term of the sequence is 25-2 = 30 therefore
the divisor cannot be greater than 30. Therefore 30 is the largest number that d
ivides each member of the sequence.