Announcing a new Wagstaff PRP - Page 2

LaurV · 2021-06-30, 07:53

Good choice! Thanks to the mod colleague who did it.

Dr Sardonicus · 2021-06-30, 13:03

Quote:

Originally Posted by LaurV

Good choice! Thanks to the mod colleague who did it.

I agree, good decision.

But we mere Supermods can't ban a blogger. That requires an Act of God. And God has heard the prayers of the afflicted, and has done an Act of Mercy.

Our kvetching kibbitzer has temporarily been quietened.

Quote:

Originally Posted by Humble Pie

Thirty days in the hole
That's what they give you now
Thirty days in the hole
Oh, yeah
Thirty days in the hole
All right, all right

paulunderwood · 2021-06-30, 13:12

I'm amazed by Sweety's SNR and the amount ensuing noise from posters, especially in this thread about Ryan's discovery of a huge PRP. Is nothing sacred?

I hope Ryan goes to 30M bits on this one. I am not saying he should do it because I say so

Dr Sardonicus · 2021-06-30, 14:01

After my tour de force

proving that (2^p + 1)/3 automatically "passes" the Rabin-Miller test to the base 2 when p > 3 is prime (and stupidly failing to point out that I had actually proven that), it occurred to me that a simple Fermat PRP test for (2^p + 1)/3 to any base is easily refined to a Rabin-Miller test in this case since (N-1)/2 is odd, and for small bases the results are predictable using pencil-and-paper calculations. We have

N = (2^15135397+1)/3

Clearly N == -1 (mod 4). For base b = 3, 5, 7, and 11, reducing the exponent mod 6, 4, 6, and 10 respectively, we find that

3*N == 3 (mod 9) so that N == 1 (mod 3), N == 1 (mod 5), N == 1 (mod 7), and N == -1 (mod 11).

Assuming N is prime, applying the law of quadratic reciprocity tells us that

(3/N) = (-1/N)(-3/N) = (-1)(N/3) = -1

(5/N) = (N/5) = +1

(7/N) = (-1/N)(-7/N) = (-1)(N/7) = -1

(11/N) = (-1/N)(-11/N) = (-1)(N/11) = (-1)(-1) = +1, so if N is prime,

3^((N-1)/2) == -1 (mod N)

5^((N-1)/2) == +1 (mod N)

7^((N-1)/2) == -1 (mod N), and

11^((N-1)/2) == +1 (mod N).

If any of these congruences fail to hold, that would prove that N is composite. Further, if any of these residues were anything other than 1 or -1 (mod N), it would give a proper factorization of N.

I am confident that ryanp et al know all this, and checked all this.

Someone who natters that "you should run a Rabin-Miller test" without even bothering to do the above paper-and-pencil calculations (and perhaps doesn't even know how), or considering that they might be addressing their quibbling to people who know all this and have almost certainly already done the calculations, should IMO thank God for His infinite mercy in not issuing a permanent ban.

ryanp · 2021-07-01, 00:20

Quote:

Originally Posted by paulunderwood

I hope Ryan goes to 30M bits on this one. I am not saying he should do it because I say so

I may, but most likely it will be later this summer.

Also, LLR with "all the switches" reports:

Code:

(2^15135397+1)/3 is BPSW and Frobenius PRP! (P = 1, Q = 3, D = -11)  Time : 51568.343 sec.

ryanp · 2021-07-12, 17:21

I've now searched the whole 13M...19M range and should finish up to 20M this week.

2021-06-30, 13:12	#14
paulunderwood Sep 2002 Database er0rr 1000000010100₂ Posts	I'm amazed by Sweety's SNR and the amount ensuing noise from posters, especially in this thread about Ryan's discovery of a huge PRP. Is nothing sacred? I hope Ryan goes to 30M bits on this one. I am not saying he should do it because I say so Last fiddled with by paulunderwood on 2021-06-30 at 13:14

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2021-06-30, 07:53	#12
LaurV Romulan Interpreter "name field" Jun 2011 Thailand 23351₈ Posts	Good choice! Thanks to the mod colleague who did it.

2021-06-30, 14:01	#15
Dr Sardonicus Feb 2017 Nowhere 3·17·113 Posts	After my tour de force proving that (2^p + 1)/3 automatically "passes" the Rabin-Miller test to the base 2 when p > 3 is prime (and stupidly failing to point out that I had actually proven that), it occurred to me that a simple Fermat PRP test for (2^p + 1)/3 to any base is easily refined to a Rabin-Miller test in this case since (N-1)/2 is odd, and for small bases the results are predictable using pencil-and-paper calculations. We have N = (2^15135397+1)/3 Clearly N == -1 (mod 4). For base b = 3, 5, 7, and 11, reducing the exponent mod 6, 4, 6, and 10 respectively, we find that 3N == 3 (mod 9) so that N == 1 (mod 3), N == 1 (mod 5), N == 1 (mod 7), and N == -1 (mod 11). Assuming N is prime, applying the law of quadratic reciprocity tells us that (3/N) = (-1/N)(-3/N) = (-1)(N/3) = -1 (5/N) = (N/5) = +1 (7/N) = (-1/N)(-7/N) = (-1)(N/7) = -1 (11/N) = (-1/N)(-11/N) = (-1)(N/11) = (-1)(-1) = +1, so if N is prime, 3^((N-1)/2) == -1 (mod N) 5^((N-1)/2) == +1 (mod N) 7^((N-1)/2) == -1 (mod N), and 11^((N-1)/2) == +1 (mod N). If any of these congruences fail to hold, that would prove that N is composite. Further, if any of these residues were anything other than 1 or -1 (mod N), it would give a proper factorization of N. I am confident that ryanp* et al know all this, and checked all this. Someone who natters that "you should run a Rabin-Miller test" without even bothering to do the above paper-and-pencil calculations (and perhaps doesn't even know how), or considering that they might be addressing their quibbling to people who know all this and have almost certainly already done the calculations, should IMO thank God for His infinite mercy in not issuing a permanent ban.

2021-07-12, 17:21	#17
ryanp Jun 2012 Boulder, CO 2⁴×23 Posts	I've now searched the whole 13M...19M range and should finish up to 20M this week.