All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Divisor Determination (Posted on 2015-04-25)
Find all possible values of an integer M such that any prime divisor of M6 - 1 divides (M3 - 1)(M2 - 1).

 No Solution Yet Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 A start? | Comment 1 of 2
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.

 Posted by Jer on 2015-04-25 16:04:04

 Search: Search body:
Forums (0)