Thread: Another Wagstaff PRP Test View Single Post
2018-05-10, 14:31   #2
Dr Sardonicus

Feb 2017
Nowhere

168A16 Posts

Quote:
 Originally Posted by paulunderwood Here are tests for Wagstaff (2^p+1)/3: If p==1 mod 6: Code: w1(p)=s=Mod(4,(2^p+1)/3);for(k=1,p-2,s=s^2-2);s==4 If p==5 mod 6: Code: w5(p)=s=Mod(4,(2^p+1)/3);for(k=1,p-1,s=s^2-2);s==-4 Can you prove or disprove either of these?
let p > 3 be a prime number, M = (2^p + 1)/3. Then M == 3 (mod 8).

Let u = Mod(x, x^2 - 4*x + 1), so that u^2 - 4*u + 1 = 0.

Let R = Z[u] = ring of algebraic integers in Q(sqrt(3)).

If p == 5 (mod 6), then M == 2 (mod 3). If M is prime, then (3/M) = +1, so MR = PP', R/P and R/P' both isomorphic to Z/MZ. Thus

u^(M-1) == 1 (mod MR).

If v = u - 1, we have norm(v) = 2, so that

v^(M-1) == 1 (mod MR) also. Now

v^2 = 2*u, so that

v^(M-1) = 2^((M-1)/2) * u^((M-1)/2). Thus

2^((M-1)/2) * u^((M-1)/2) == 1 (mod MR). Now (2/M) = -1, so we have

u^((M-1)/2) == -1 (mod MR). Cubing, we obtain

u^(2^(p-1) - 1) == -1 (mod MR), or

u^(2^(p-1)) == -u (mod MR),

which validates Paul's test as a necessary condition that (2^p + 1)/3 be prime, when p == 5 (mod 6).

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If p == 1 (mod 6), then M == 1 (mod 3). If M is prime, then (3/M) = -1, so MR is a prime ideal, and R/MR = GF(M^2). In this case, the Frobenius automorphism gives us that, for any r in R, if r' is the algebraic conjugate of r,

r^M == r' (mod MR), so that

r^(M+1) == r*r' = norm(r) (mod M). In particular,

u^(M+1) == 1 (mod MR) and

v^(M+1) == -2 (mod MR). Proceeding as above, we obtain

-2 == 2^((M+1)/2) * u^((M+1)/2).

Now 2^((M+1)/2) = 2*(2^((M-1)/2) == -2 (mod M), since (2/M) = -1. Therefore

u^((M+1)/2) == +1 (mod MR). This tells us that

u^((M+1)/4) == +1 or -1 (mod MR). If +1 is correct, we obtain

u^(2^(p-2)) == u^(-1) (mod MR), which would validate Paul's test as a necessary condition that (2^p + 1)/3 be prime, when p == 1 (mod 6).

Alas, I have so far been unable to determine in general whether it is +1 or -1.

I note that in this case, M == 19 (mod 24). I looked at

u^((q+1)/4) (mod qR) for q prime, q == 19 (mod 24)

and found that

u^((q+1)/4) == +1 (mod qR) about half the time, and

u^((q+1)/4) == -1 (mod qR) about half the time.

The smallest q == 19 (mod 24) for which

u^((q+1)/4) == -1 (mod qR)

is q = 67.

And that's as far as I've gotten.