20060101, 06:20  #1 
May 2004
2^{2}×79 Posts 
Mersenne Prime Factors of v.large numbers
For the above refer to "Minimum Universal Exponent Generalisation of
Fermat's Theorem" on site: www.crorepatibaniye.com/failurefunctions A.K. Devaraj (dkandadai@yahoo.com) 
20060102, 17:03  #2 
Aug 2002
Buenos Aires, Argentina
5D4_{16} Posts 
In your web site you wrote:
is the minimum universal exponent of m (m belongs to N). What do you mean by minimum universal exponent? I suppose that by N you refer to N, the set of natural numbers. 
20060104, 03:54  #3  
May 2004
100111100_{2} Posts 
Quote:
5 is the minimum universal exponent (w.r.t base 2) i.e. (2^5)  1 =31 and 5 is the minimum exp such that 2^n  1 is congruent to zero (mod 31). Devaraj 

20060104, 04:05  #4  
May 2004
316_{10} Posts 
Quote:
31, a Mersenne Prime , for any x ending with 1 or 6. b) 127, another Mersene prime, is an impossible factor of this function i.e. it cannot be a factor of (2^x) + 29, no matter how large x is. Devaraj 

20060104, 06:58  #5  
"Richard B. Woods"
Aug 2002
Wisconsin USA
7692_{10} Posts 
Quote:
(2^3)  1 =7 and 3 is the minimum (positive) exponent such that 2^n  1 is congruent to zero (mod 7). (2^2)  1 =3 and 2 is the minimum (positive) exponent such that 2^n  1 is congruent to zero (mod 3). 

20060104, 09:36  #6  
Jun 2005
Near Beetlegeuse
2^{2}×97 Posts 
Quote:
I think this is a consequence of the corollary he mentioned in his previous post Quote:


20060104, 22:44  #7 
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}·3·641 Posts 
Okay. Thanks.

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