 mersenneforum.org Possible proof of Reix' conjecture (Wagstaff primes, plus some issues)
 Register FAQ Search Today's Posts Mark Forums Read 2020-09-10, 16:29 #1 tetramur   "unknown" Jan 2019 anywhere 17 Posts 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). Proof: 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 8t^2-2t-1<0 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   2021-03-20, 18:18 #2 T.Rex   Feb 2004 France 91810 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   2021-07-11, 09:31   #3
tetramur

"unknown"
Jan 2019
anywhere

17 Posts Quote:
 Originally Posted by T.Rex 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
Thanks for your feedback. Anyway, it'll be interesting to prove Conjecture 1. Maybe one day I'll post the proof here.   2021-09-20, 18:58   #4
kijinSeija

Mar 2021
France

22 Posts Quote:
 Originally Posted by tetramur Thanks for your feedback. Anyway, it'll be interesting to prove Conjecture 1. Maybe one day I'll post the proof here.
(3+sqrt(-7/4))*(3-sqrt(-7/4)) = 43/4 right ? I'm not sure about that   2021-09-20, 19:33   #5
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

100101011001102 Posts Quote:
 Originally Posted by kijinSeija (3+sqrt(-7/4))*(3-sqrt(-7/4)) = 43/4 right ? I'm not sure about that
Use (a-b)(a+b) = a2 - b2   2021-09-21, 13:34   #6
Dr Sardonicus

Feb 2017
Nowhere

116168 Posts Quote:
 Originally Posted by tetramur 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).
There is no such thing as "the rationals mod q." The rational number 1/q is not defined "mod q."

The residue ring Z/qZ of the integers mod q is a field, the finite field Fq of q elements. In the field K = Q(sqrt(-7)), the ring of algebraic integers is R = Z[(1 + sqrt(-7))/2].

If -7 is a quadratic non-residue (mod q) [that is, if q == 3, 5, or 13 (mod 14)], then the residue ring R/qR is the finite field of q2 elements.

If -7 is a quadratic residue (mod q) [that is, if q == 1, 9, or 11 (mod 14)] the residue ring R/qR is not a field, but is the direct product Fq x Fq of two copies of Fq.

An example of the latter case where (2^p + 1)/3 is composite is p = 53, for which q = 107 == 9 (mod 14).  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post GP2 Wagstaff PRP Search 414 2020-12-27 08:11 Tony Reix Wagstaff PRP Search 7 2013-10-10 01:23 davieddy Miscellaneous Math 209 2011-01-23 23:50 AntonVrba Math 96 2009-02-25 10:37

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