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(p1) − 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(p1) − S0 = 0 (mod Wp). Then
w^2^(p1) + v^2^(p1)  w  v = k*Wp
for some integer k, so
w^2^(p1) = k*Wp  v^2^(p1) + w + v
w^2^p = k*Wp*w^2^(p1)  1 + w^(2^(p1)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* <= X1=q^42*q^3+q^21.
[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 q1 possibilities for b, because 0 has no inverse in X. This gives q(q1) possibilities for one rational and (q(q1))^2 for two rationals, equal to q^42*q^3+q^2 elements.]
Now Wp = 0 (mod q) and w is in X, so k*Wp*w^2^(p1) = 0 in X. Then by (1),
w^2^p = 1 + w^(2^(p1)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^p1, 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^p1 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^42*q^3+q^21 < 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^22t1<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 VrbaGerbicz theorem) is actually true (I think) and we have an efficient primality test for Wagstaff numbers  deterministic, polynomial and unconditional, similar to LucasLehmer test for Mersenne numbers.
Last fiddled with by tetramur on 20200910 at 16:38
