20040723, 08:15  #1 
May 2004
474_{8} Posts 
Fermat Numbers
Nonfactors of Fermat Numbers:
1) No prime number generated by 2 exp (2n + 1) + 1 can be a factor of any Fermat Number 2 ) prime numbers not generated by 2exp n + 1 also cannot be factors Any other sets? ( to be contd) Devaraj 
20040723, 16:13  #2  
Bronze Medalist
Jan 2004
Mumbai,India
2052_{10} Posts 
Fermat Numbers
Quote:
Mally 

20040724, 00:00  #3 
Sep 2002
2×331 Posts 
2exp n + 1 is that 2^n + 1 ? ( one plus 2 to the nth power)
What is n ? Is it any integer large enough to allow 2^n + 1 to be a factor ? Is 2 ) equivalent to: only prime numbers 2^n + 1 can be factors of fermat numbers ? IE Binary numbers of the form 101 1001 10001 
20040724, 05:12  #4 
May 2004
2^{2}×79 Posts 
Fermat Numbers
Reply to Moderator: yes to the first query.
Also n is any positive integer Reply to "Mally": These are corallaries of my theorem: "A Generalisation of Fermat's Theorem" on my site: www.crorepatibaniye.com/failurefunctions Devaraj 
20040725, 01:45  #5 
Sep 2002
2×331 Posts 
concerning point 2)
if I correctly understand what you wrote, then it isn't correct. Counter example Fermat5 = 2^(2^5) + 1 = 2^32 + 1 = 4294967297 = 100000000000000000000000000000001 (binary) the possible factors are of the form k*(2^7) + 1 with k >= 0 with k = 0 factor = 1 with k = 5 factor = 5*(2^7) + 1 = 5*128 + 1 = 641 = 1010000001 641 is prime, is a factor of Fermat5, isn't 2^n + 1 
20040726, 04:28  #6 
May 2004
2^{2}×79 Posts 
Fermat Numbers
To the moderator:
641 is a factor of F5 i.e. 2^2^5 + 1; it is not a factor of 2^(2n + 1) + 1. Hope this clarifies the matter. Devaraj 
20040726, 13:27  #7  
Nov 2003
2^{2}·5·373 Posts 
Quote:
Allow me to clarify this matter. I mean no disrespect, but the original poster would greatly benefit from actually READING about this subject. Any primitive prime factor of 2^k+1, must equal 1 mod 2k. Period. Primitive means that it does not divide 2^m+1 for any m < k. e.g. 11 is a primitive prime factor of 2^5+1. It is also a prime factor of (say) 2^15+1, but it is not a primitive prime factor. For Fermat numbers, when k = 2^n (say) we can say a little bit more because 2 is a primitive root. In this case, we have that any prime factor is 1 mod 4k. Any two different Fermat numbers are also relatively prime. Now can we put the random speculations to rest???? 

20040727, 04:12  #8 
May 2004
2^{2}×79 Posts 
Fermat Numbers
Thank you for further clarification, Dr. Silverman.,
Devaraj Last fiddled with by devarajkandadai on 20040727 at 04:13 
20040727, 12:27  #9  
Nov 2003
7460_{10} Posts 
Quote:
"because 2 is a primitive root. " should clearly have been: because 2 is a quadratic residue. For those who are interested, here is a full proof. Assume 2^2^n + 1 = 0 mod p Then 2^2^n = 1 mod p 2^2^(n+1) = 1 mod p By Lagrange's Thm, the order of any element must divide the order of the group. The order of the group is p1, hence 2^(n+1)  p  1 and therefore p = k* 2^(n+1) + 1 for some k. But we can say more. Note that for n > 1, we also have p = 1 mod 8. By quadratic reciprocity 2 is a quadratic residue. From Euler's criteria we then obtain: 2^ ((p1)/2) = 1 mod p and hence (2^(n+1))  (p1)/2. This last result yields p = k^2^(n+2) + 1. Euler's criterion says that a^((p1)/2) = 1 or 1 mod p depending on whether a is (resp.) a residue or nonresidue. 

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