Go Back > Prime Search Projects > Wagstaff PRP Search

Thread Tools
Old 2020-09-10, 16:29   #1
Jan 2019

D16 Posts
Post Possible proof of Reix' conjecture (Wagstaff primes, plus some issues)

I claim to have proved the Reix' conjecture (2007), part "if":

Theorem 1.1. Let p > 3 prime and for the sequence S0 = 3/2, Sk+1 = Sk^2 − 2 it is true that S(p-1) − S0 is divisible by W(p), then W(p) = (2^p+1)/3 is also prime (Wagstaff prime).

Let w = 3+√-7/4 and v = 3-√-7/4. Then it is proved by induction then Sk = w^2^k + v^2^k.

Suppose S(p-1) − S0 = 0 (mod Wp). Then
w^2^(p-1) + v^2^(p-1) - w - v = k*Wp
for some integer k, so
w^2^(p-1) = k*Wp - v^2^(p-1) + w + v
w^2^p = k*Wp*w^2^(p-1) - 1 + w^(2^(p-1)-1)*(w^2+1) (1)
[w*v = 1, it can be easily proved: 9/16 - (-7)/16 = 9/16 + 7/16 = 1]

We are looking for contradiction - let Wp be composite and q be the smallest prime factor of Wp. Wagstaff numbers are odd, so q > 2.

Let Q_q be the rationals modulo q, and let X = {a+b √-7} where a,b are in Q_q. Multiplication in X is defined as
(a+b√-7)(c+d√-7) = [(ac - 7bd) mod q] + √-7 [(ad+bc) mod q]
Since q > 2, it follows that w and v are in X. The subset of elements in X with inverses forms a group, which is denoted by X* and has size |X*|. One element in X that does not have an inverse is 0, so |X*| <= |X|-1=q^4-2*q^3+q^2-1.
[Why is it? Because X contains pair of rationals modulo q, and suppose we have rational a/b mod q. We have q possibilities for a and q-1 possibilities for b, because 0 has no inverse in X. This gives q(q-1) possibilities for one rational and (q(q-1))^2 for two rationals, equal to q^4-2*q^3+q^2 elements.]
Now Wp = 0 (mod q) and w is in X, so k*Wp*w^2^(p-1) = 0 in X. Then by (1),
w^2^p = -1 + w^(2^(p-1)-1)*(w^2+1)

I want to find order of w in X and I conjecture it to be exactly 2^(2*p). [I couldn't resolve this when I was working for a proof.] Why is it? If we look to similar process to 2^p-1, w = 2+√3, v = 2-√3, we have equality w^2^p = 1, order is equal to 2^p, but it is the first power of 2 to divide 2^p-1 with remainder 1. Similarly, 2^(2*p) is the first power of 2 to divide Wp with remainder 1, and I conjecture that it is the true order.

The order of an element is at most the order (size) of the group, so
2^(2*p) <= |X*| <= q^4-2*q^3+q^2-1 < q^4.
But q is the smallest prime factor of the composite Wp, so
q^4 <= ((2^p+1)/3)^2.
This yields the contradiction:
9*2^2p < 2^2p + 2^(p+1) + 1
8*2^2p - 2*2^p - 1 < 0
2^p = t
D = 4+4*8=36 = 6^2
t1,2 = 2+-6/16
t1 > -1/4
t2 < 1/2, 2^p < 1/2, p < -1
Therefore, Wp is prime.

So, I think it is almost proven, but there is one issue.
Conjecture 1. Let p be prime p > 3, q be the smallest divisor of Wp = (2^p+1)/3 and both a, b be rationals mod q, then the order of the element w = 3+√-7/4 in the field X of {a+b √-7} is equal to 2^(2*p).

If both conjecture and proof turn out to be true, then converse of Reix' conjecture (that is, converse of Vrba-Gerbicz theorem) is actually true (I think) and we have an efficient primality test for Wagstaff numbers - deterministic, polynomial and unconditional, similar to Lucas-Lehmer test for Mersenne numbers.

Last fiddled with by tetramur on 2020-09-10 at 16:38
tetramur is offline   Reply With Quote
Old 2021-03-20, 18:18   #2
T.Rex's Avatar
Feb 2004

22×229 Posts

Hi tetramur,
I've just seen your post. I need to refresh my few Maths skills before saying anything. And probably that I'll not be able to say if it is correct or not.
Anyway, I'm happy to see that people are still trying to prove this. Thanks.
Tony Reix
T.Rex is offline   Reply With Quote

Thread Tools

Similar Threads
Thread Thread Starter Forum Replies Last Post
Status of Wagstaff testing? and testing Mersenne primes for Wagstaff-ness GP2 Wagstaff PRP Search 414 2020-12-27 08:11
500€ Reward for a proof for the Wagstaff primality test conjecture Tony Reix Wagstaff PRP Search 7 2013-10-10 01:23
Wagstaff Conjecture davieddy Miscellaneous Math 209 2011-01-23 23:50
PRIMALITY PROOF for Wagstaff numbers! AntonVrba Math 96 2009-02-25 10:37

All times are UTC. The time now is 16:11.

Mon Jun 14 16:11:00 UTC 2021 up 17 days, 13:58, 0 users, load averages: 1.45, 1.54, 1.46

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.