20140103, 19:33  #100 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
22165_{8} Posts 
Looks new. You want to report it to W.Keller:
http://www.prothsearch.net/GFNfacs.html You were in the right zone, just above the previous search limit: http://www.prothsearch.net/GFNsrch.txt 
20140103, 19:39  #101 
Apr 2010
Over the rainbow
2×5×11×23 Posts 
I did, it is a copy of the mail I just sent. the last time I checked it, the search limit was different.... oh well, that will remove some test for the following exponent. Onward to F6152.
Last fiddled with by firejuggler on 20140103 at 19:41 
20140114, 21:13  #104 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{2}×17×61 Posts 
Of course. pfgw gxo q"insert number here"
(in case of the number above, pfgw gxo q"9*2^3497442+1", but for validation that it is a GF(10), pfgw gos10 q"9*2^3497442+1" would work. Add a1 for a different FFT size, to doublecheck.) 
20140121, 00:06  #105  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10010001110101_{2} Posts 
Here are some thoughts (and a placeholder for a picture) for the specific Proth prime numbers (k=3). There are not original, just mental floss.
This is related to the latest large Proth prime 3*2^10829346+1 To start with, it divides GF(10829343,3), GF(10829345,5), GF(10829344,8) (the algebraic cofactor of it), and GF(10829345,11). Quote:
Let p=3*2^n+1, calculations be mod p, and consider a sequence s(0)=b>1, s(k)=s(k1)^2, 1<=k<=n. The final residues of the n squarings (in any base b) are cube roots of 1 (mod p), because after cubing we get the Fermat’s little theorem b^(p1)=1 (mod p). If the final residue is 1, then the paths leading to it are only through square roots of 1, which are 1 and 1, so after a few steps back we have a GF divisor (because the initial value was b>1, not all residues are equal to 1, so there exists an iteration with value 1). The other two roots are L and M, such that L+M=1, L·M=1, L^2=M, and M^2=L (mod p). For p=3*2^n+1 and n even, they happen to be L=±3(2^(n1)2^(n/21)), M=±3(2^(n1)+2^(n/21)) and are related to Aurifeuillean factors of 2^(2n2). (+ signs are as in Robinson’s). They are coincidentally the solutions to the length2 cycle question: which values x lead to length2 cycles in the sequence s(n)? This requires x^4=x, which (with x≠0) is equivalent to x^3=1, and therefore (x1)*(x^2+x+1)=0 (mod p). x=1, or x=L, or x=M. These roots x form some of the possible entrances into terminal cycles of squaring (in addition to trivial 1 > 1 > 1...): L > M > L > M > L... and M > L >M > L >M... Indeed, all these paths in different combinations are observed for b=2, 3, 5, 7, 11. For a composite base c GFdivisor test, the residues s(m,b_i) (where b_i are the prime factorization of c) have to be multiplied and the product to be 1 for pGF(m,c). For example, b=5 and b=11 have final values (for n2<=m<= n), {R, 1, 1} and {R,1,1} respectively, where R is the smaller (“positive”) sqrt(1) (mod(p)). So, p divides xGF(n2,5,11) and some GF(•,55). For c=14=2·7, p divides GF(n2,14), because L·M=1. OTOH, products of L and M values are again values L, M or 1. Square roots of L are ±M, and likewise for M. Combinatorial products of 1, –L or –M with some L, M or 1s, produce the desired 1, for example, p divides GF(n1,70), because s(n1,70)=1·L·M=1. p also divides GF(n2,8), because L^{3}=1. Some xGFs can be calculated for (1<a<b<=12), e.g. p divides xGF(n4,7,4), xGF(n3,7,12), xGF(n2,3,8), xGF(n2,9,8), xGF(n2,5,11), xGF(n1,3,5), xGF(n1,8,5), xGF(n1,9,5), xGF(n1,3,11), xGF(n1,8,11), xGF(n1,9,11), etc. See attached table. Last fiddled with by Batalov on 20140121 at 07:52 Reason: fixed typos and errors 

20140121, 06:19  #106 
Jan 2005
2·31 Posts 
I believe that some time back in the 1980's or early 90's Hiromi Suyama noticed that Morehead's Theorem (as quoted by Gallot above) extends to any prime of the form 3*(k^2)*2^n + 1 with n even and k odd, with essentially the same proof. He published this in the AMS Abstracts, but unfortunately there are no online archives available that I could find.

20140121, 07:13  #107 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{2}·17·61 Posts 
It is interesting that out of 48 known Proth primes with k=3 and n>1, 40 divide GF(•,3) (including the last one).
The 48 primes are in two classes n mod 3:
The 48 primes are in two classes n mod 2:
*There is probably a rather simple proof, too? A la Morehead/Suyama? Last fiddled with by Batalov on 20140121 at 09:45 
20140121, 15:16  #108  
Jan 2005
62_{10} Posts 
Quote:


20140121, 19:36  #109 
Jan 2005
2×31 Posts 
I see my error; the condition wasn't k odd but k=+/1 mod 6. Then the unique representation of a prime N = 3*(k^2)*2^n+1 (n even) in the form A^2 + 3*B^2 will have A=1 and B=k*2^(n/2). When k is not divisible by 3, 2 is not a cubic residue and N cannot be a Fermat factor.

20140121, 19:38  #110 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3^{2}×17×61 Posts 
Yves shared a reference to the Calvo (2000) paper. Indeed, his theorem 2.1 explains the special case kn and more.

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