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Divisor Determination (Posted on 2015-04-25) Difficulty: 3 of 5
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    
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Some Thoughts re: A start? Comment 2 of 2 |
(In reply to A start? by Jer)

Jer,

You did the hard part:

1. Clearly, m never divides m^2-m+1

2. Note that m^2-m+1 =(m(m-1))+1. Since (m-1) divides m^2-m, it never divides m^2-m+1 either.

3. But we also have, for any m, m^2-m+1 = (m-1)^2+(m-1)+1:       
(m+1) = (m-1)+2       
Let (m-1)  = n:       
(n^2+n+1)/(n+2) = (n-1)+3/(n+2); integral iff n = {1, -1, -3}     

4. m^2-m+1+2m = m^2+m+1; as you point out, clearly, these two expressions have no common divisors greater than 1.       
       
So the only candidate is n = 1, i.e. m = 2: 2^6-1 = 63 = 3^2x7; (2^3-1)(2^2-1) = 3*7, satisfying the condition.


  Posted by broll on 2015-04-25 23:49:52
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