Well you can start by factoring
m^6-1=(m+1)(m-1)(m^2+m+1)(m^2-m+1)
and
(m^3-1)(m^2-1)=(m+1)(m-1)^2(m^2+m+1)
So all but one of the factors of m^6-1 is guaranteed to divide (m^3-1)(m^2-1)
the only one left to worry about is (m^2-m+1)
So the problem can be reworded as:
Find all possible values of an integer M such that any prime divisor of (M^2-M+1) divides at least one of: (M+1), (M - 1), (M^2+M+1).
M=2 is the only solution I've found (not that I searched far).
It seems we can rule out (M^2+M+1) because the sequence
it generates is just one term ahead and every 3rd term has a factor of 3, every 7th has a factor of 7 etc. So they have no common factors.
Which leaves just (M+1) and (M-1).
I'm stuck here.
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Posted by Jer
on 2015-04-25 16:04:04 |
A start?
