2020-09-10, 16:29 | #1 |
"unknown"
Jan 2019
anywhere
15_{8} 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 |
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