I'm looking for a complete proof for the following conjecture: Conjecture 1 Mq = 2q â 1 . S0 = 32 + 1/32 ... Here after, (a | b) is the Legendre symbol. All references ...
A LLT-like test for Mersenne numbers, based on cycles of the Digraph under x2 − 2 modulo a Mersenne prime Tony Reix 2007, 10th of May - Updated 2008, 8th of September. I Version 0.4 J
I’m looking for a complete proof for the following conjecture: Conjecture 1 Mq = 2q − 1 . S0 = 32 + 1/32 , Si+1 = Si2 − 2 (mod Mq ) Mq is a prime iff Sq−1 ≡ S0 (mod Mq ) And we have:
Qq−1 1
Si ≡ 1 (mod Mq ) when Mq is prime
I perfectly know that this test cannot speed up the proof of primality of a Mersenne number. But the method used for proving it could be used to prove that a Mersenne is not prime faster than the classical LLT. Or it could be used for other numbers for which no LLT test does exist. This conjecture makes use of the properties of the cycles of length q − 1 that appear in the Digraph under x2 − 2 modulo a Mersenne prime ; though the LLT makes use of the properties of the tree of the same Digraph. It has been checked with huge values of q. Thanks to the help of H.C. Williams, who suggested me to use the little Fermat theorem, the first part of the conjecture has been proved. Now, how can we prove the converse ?
1
Definitions
The first part of the proof makes use of the Lucas Sequence method, as described in many papers and books, like ”The Little Book of Bigger Primes” ´ by Paulo Ribenboim or like ”Edouard Lucas and Primality Testing” by Hugh C. Williams. Here after, (a | b) is the Legendre symbol. All references to theorems apply to properties of Lucas Sequences as given by P. Ribenboim in his book ”The Litlle Book of Bigger Primes” in 2.IV pages 44-etc . Since q is prime, we have: Mq ≡ 1 (mod 6q) . Let: β = 32 and α e ≡ 1/β (mod Mq ) . Since Mq is prime, α e is the only integer such that 0 < α e < Mq . 1
There are an infinity of α > Mq such that α ≡ α e (mod Mq ). Here below, we explain how to compute α e and some α . When q ≡ 1 (mod 3) and since q is odd, we have: q ≡ 1 (mod 6) . Thus Mq = 2q − 1 = 26k+1 = 2(23 )2k − 1 ≡ 1 (mod 9) . Thus 8Mq + 1 ≡ 0 8M +1 (mod 9) = 9e α and α e = 9q ≡ 1/9 (mod Mq ) . When q ≡ 2 (mod 3) and since q is odd, we have: q ≡ 5 (mod 6) . Thus Mq = 2q − 1 = 26k+5 = 32(23 )2k − 1 ≡ 4 (mod 9) . Thus 2Mq + 1 ≡ 0 2M +1 (mod 9) = 9e α and α e = 9q ≡ 1/9 (mod Mq ) . With q > 5, we always have: β < α e
Fermat theorem, the first part of the conjecture has been proved. Now, how can ... All references to theorems apply to properties of Lucas Sequences as given.
All references to theorems apply to properties of Lucas Sequences as given ... For each pair (a, b) of positive integers such that: Mq = F1F2 , there exist.
Conjecture 1 (Mersenne numbers) is mine, based on my work on the use of the Cycles of the Digraph under x2 − 2 modulo a Mersenne prime for primality ...
G1Ã4+0[4] mod 5 = G1Ã16+0[16] mod 17 = G1Ã256+0[256] mod 257 = 0. G2Ã4+0[4] mod 5 = G2Ã16+0[16] mod 17 = G2Ã256+0[256] mod 257 = 2. G2Ã4+1[4] ...
order to try proving a faster Primality test of Fermat numbers. Though these properties are detailed in the ... V 2 n = 4(2U2 n + (â1)n) or V 2 n â 8U2 n = (â1)n4. 2 ...
3 that divides the order of 3 modulo a Mersenne prime is 2 ? or not... So, my guess is: yes. But maybe the STRONG âlaw of SMALL numbersâ will apply there ...
A property dealing with the order of 3 modulo a Mersenne prime. Tony Reix ([email protected]). ZetaX (AOPS forum). 2008, 8th of March. In May of 2006, ...
1. Definition of the LLT numbers. Let's say: L(x) = x2 − 2 , L1 = L , Lm = L◦Lm−1 = L◦L◦L... ◦ L. ... C− m the sum of the negative coefficients of the polynomial Lm(x) . ... −2 (mod Fn) for: m > 1 ( LLT.5). C+ m ≡ 3 (mod 10) for: m ≥ 1 ( LLT.6). 2n−1
based on iteration of the quadratic maps x â x2 and x â x2 â2 over a finite .... x does not necessary exist in Fp, but, since it is a quadratic equation, for.
c a c b a c. c c a d a a d. a c c b c a b. a a c d a c b a. Denise Vella-Chemla. Goldbach's conjecture, 4 letters language, variables and invariants. May 2014. 5 / 23 ...
A very simple property of Mersenne numbers. Tony Reix. 2009, 11th of April. â· Version 0.1 â. Let show that: 2q â 1 = 1 + 6(1 + 22 + 24 + 26 + ... + 22qâ1. 2. ) .
This paper provides a proof of a LLT-like test for Fermat numbers, based .... Proof: Since p is a prime, and by Fermat little theorem, we have: 2pâ1 â¡ 1 (mod p).
The proof is based on the chapters 4 (The Lucas Functions) and 8.4 (The ... Proof: Since p is a prime, and by Fermat little theorem, we have: 2pâ1 â¡ 1 (mod p).
and Si = 2S 2 iâ1 â 1 for i = 1,2,3,... q â 2 . The proof is based on the chapters 4 (The Lucas Functions) and 8.4 (The. Lehmer Functions) of the book âÃdouard ...
serie exhibits the same kind of properties than the Fermat numbers. an = 4n + 2n .... The proof of (FLS.7) by Saouter provides nearly all we need for proving ... Since by Fermat little theorem we have: 2pâ1 â¡ 1 (mod p), then Ï also divides pâ1
This paper provides the proof of two new primality tests for Fermat numbers, .... Since Vn = 3n + 1 as shown in page 1, and by the previous theorem, we have: Fn ...
migrated separately to the U.S. while still teenagers. By the time they met, married and produced a son, the process of Americanization was com- plete and Tony ...
