mersenneforum.org  

Go Back   mersenneforum.org > Search Forums

Showing results 1 to 25 of 33
Search took 0.01 seconds.
Search: Posts Made By: Tony Reix
Forum: Wagstaff PRP Search 2013-10-09, 19:34
Replies: 7
Views: 8,531
Posted By Tony Reix
Review of the Conjectures

I've asked Hugh Williams his opinion about the conjectures. He accepted and he first wanted to look at the conjectures and the proofs of necessity. He is not really sure that they are correct or...
Forum: Wagstaff PRP Search 2013-10-06, 16:08
Replies: 7
Views: 8,531
Posted By Tony Reix
500€ Reward for a proof for the Wagstaff primality test conjecture

Hi,

I've decided to offer a 500€ reward for the first guy/lady who will be able to provide a proof for the conjecture that Anton Vrba and myself have found, years ago, about proving the primality...
Forum: News 2013-01-28, 20:24
Replies: 571
Views: 71,978
Posted By Tony Reix
Waowww !

Wonderful ! A new candidate Mersenne prime ! Nice work, guys !

I do not have access to any big Bull computer for checking this new candidate. But others should have access to big machines, for...
Forum: Math 2003-03-04, 17:03
Replies: 7
Views: 751
Posted By Tony Reix
Mq = 7 (mod 120) for q = 4k + 3

We have
2^4 * 2^3 = 16 * 8 = 128 = 8 (mod 120)
Thus
2^(4k+3) = 2^(4k) * 8 = 2^(4*(k-1)) * [ 16 * 8 ] = 2^(4*(k-1)) * 8 (mod 120) = 2^(4*(k-2)) * 8 = ... = 1 * 8 = 8 (mod 120)
And finaly
M(4k+3)...
Forum: Math 2003-03-03, 09:35
Replies: 7
Views: 751
Posted By Tony Reix
More data. Plus some conclusion.

Here are more data (Mq = 7 (mod 120)) :



These data show:
- if Mq is prime, it seems there is only 1 pair (a,b) that is solution of each above form.
- If Mq is composite, it seems there is 1...
Forum: Math 2003-02-28, 16:55
Replies: 7
Views: 751
Posted By Tony Reix
Mq=x^2+ky^2 with (x,y) unic for all k <==> Mq prime ??

After some more computation, it seems that for each q prime (that I've tested: 5, 7, 11, 13, 17, 19, 23), there are many k such that Mq=x^2+ky^2 has (at least) a (x,y) solution.
But it seems that...
Forum: Math 2003-02-28, 13:09
Replies: 7
Views: 751
Posted By Tony Reix
M19 too !

I forgot M19 :exclaim: :

Which confirms the 3 conjectures :D .

We also have the noticeable relation-ships :surprised: :

But this kind of relation-ship doesn't appear with the others (M7,...
Forum: Math 2003-02-28, 11:21
Replies: 7
Views: 751
Posted By Tony Reix
More q such that Mq = 7 (mod 120).

The following prime q such that Mq = 7 (mod 120) are:
Forum: Math 2003-02-28, 11:12
Replies: 7
Views: 751
Posted By Tony Reix
x^2 + 5*y^2

I found the previous data while looking at the theorems:


If we look at small factors of M11 and M23, we can see :


Then I saw that some Mq are of the form: 20k+7 and I tried with the form:...
Forum: Math 2003-02-28, 11:02
Replies: 7
Views: 751
Posted By Tony Reix
Mq = 7 (mod 120) ==> Mq = (2x)^2 + q*(3y)^2 ??

We have the following data :D :

Is this enough for building the conjecture :question: :


We also have the data :D :

Is this enough for building the conjectures :question: :
Forum: Math 2003-02-27, 13:20
Replies: 22
Views: 1,716
Posted By Tony Reix
Philmoore's algorithm with GP/Pari.

Here is the GP code for Philmoore's algorithm (for Mersenne primes), tested with gp v 2.1.1 .
It's quite slow! (I'm not using GMP kernel).
[code:1]
tony(q) =
{
p=2^q-1;
b=3^(2^(q-2)) % p;...
Forum: Math 2003-02-26, 14:26
Replies: 22
Views: 1,716
Posted By Tony Reix
(a/c)^2 + 3*(b/d)^2 = Mq

In the previous discussions, we have considered A^2+3*B^2 = Mq (I) where A and B are integers.
If we now consider A and B as fractions, it appears that Mersenne primes seem to have many solutions to...
Forum: Math 2003-02-20, 10:14
Replies: 22
Views: 1,716
Posted By Tony Reix
New properties about 8n-1 and 6n+1 numbers.

Hi,
In an old French book about Arithmetic, I've found some theorems that apply to Mersenne numbers.
Some of them are close to things we already know about Mersenne numbers.

1) Prime numbers of...
Forum: Math 2003-02-14, 16:57
Replies: 22
Views: 1,716
Posted By Tony Reix
A relationship between solutions of M37, and of M67.

M37 has 2 (a,b) solutions
(185341,1119)
(142357,45695)
If we find the factors of a1-a2 and b2-b1, we find
(185341-142357) = 2^3 * 3^3 * 199
(45695-1119) = 2^5 * 7 * 199

199 !

Same...
Forum: Math 2003-02-14, 16:49
Replies: 22
Views: 1,716
Posted By Tony Reix
Mistake about M27

I made a mistake about M27.
There are 4 solutions
(a,b) = (5249,943)
(A,B) = (3350,6403) (7958,4861) (9470,3853)
Forum: Math 2003-02-14, 08:53
Replies: 22
Views: 1,716
Posted By Tony Reix
(Easy!) proof of A=2a .

Here is the (easy!) proof of A=2a for (q,Mq) both primes.

Mq = 2^q - 1 = A^2 + 3*B^2 .
Since Mq is odd, we have either (A odd, B even) or (A even, B odd).
Let say B=2b and A=2a+1.
Then...
Forum: Math 2003-02-13, 17:49
Replies: 22
Views: 1,716
Posted By Tony Reix
a >= b ?

Thanks for all the information. :D

I've noticed that, for q and Mq both primes, in the formula (II) : Mq = (2*a)^2 + 3*(3*b)^2 , a is always greater than or equal to b. That's true for:...
Forum: Math 2003-02-12, 17:30
Replies: 22
Views: 1,716
Posted By Tony Reix
M29 = 2^29-1 = 233*1103*2089 233 = 2 (mod 3) ...

M29 = 2^29-1 = 233*1103*2089
233 = 2 (mod 3)
1103 = 2 (mod 3)
2089 = 1 (mod 3)
And I found no (A^2+3*B^2) representation for M29.

M67 = 2^67-1 = 193707721 * 761838257287
193707721 = 1...
Forum: Math 2003-02-12, 09:13
Replies: 22
Views: 1,716
Posted By Tony Reix
We are making progress !

Hello, thanks for looking at this conjecture !


Good ! :D
Would you mind providing the value of (a,b) you found for these 4 Mersenne primes ?

About M27, which is not prime, I missed the pair...
Forum: Math 2003-02-10, 17:16
Replies: 22
Views: 1,716
Posted By Tony Reix
Some clarifications

Since I think my previous note may be unclear, here are some clarifications.

When q and Mq are primes
1) it is certain that Mq = A^2 + 3*B^2, and (A,B) is unic.
2) it's easy to prove that A is...
Forum: Math 2003-02-10, 10:00
Replies: 22
Views: 1,716
Posted By Tony Reix
Mq prime, then: Mq = A^2 + 3B^2 with A=2a. (and B=3b ??)

There is a theorem (proven by Euler and used for proving the case n=3 of Fermat's Last Theorem), saying that:
if p is of the form: p = 1 + 3n and is prime, then there is a unic pair (A,B) such that:...
Forum: Math 2002-11-19, 08:15
Replies: 10
Views: 2,262
Posted By Tony Reix
Hi, After some more search about computing...

Hi,
After some more search about computing inverses, I've found:
[code:1]
We want to compute 1/p (mod q).
0 &lt; p &lt; q
If: p * 1/p ~= 1 (mod q) (where 1/p is the inverse of p)
with: 0 &lt; 1/p &lt; q ...
Forum: Math 2002-11-15, 15:01
Replies: 10
Views: 2,262
Posted By Tony Reix
Proof for 1/8 formula.

I forgot to say that, for n = 8, the table is:
[code:1]
1 2 3 4 5 6 7
7 0 5 0 3 0 1
[/code:1]

Since, with q prime, (q mod 8) is odd, then it's easy to say that the formula I...
Forum: Math 2002-11-15, 14:52
Replies: 10
Views: 2,262
Posted By Tony Reix
Pre-compute k for finding 1/n (mod q)

Hi ewmayer,
Thanks for explaining how to compute n^(-p) (mod q).
I think I read that long time ago, but I forgot ...
Your solution for 2(^-3) (mod q) is simpler than my formula.
But, for n prime,...
Forum: Math 2002-11-13, 17:19
Replies: 10
Views: 2,262
Posted By Tony Reix
Hi Bruce, About my formula for 1/8 1/8...

Hi Bruce,

About my formula for 1/8
1/8 (mod q) = [(8 - (q mod 8 ))*q + 1] / 8
I'm afraid I can't remember how I found it and if I proved it ...
Just now, I've experimented with all 9999 first...
Showing results 1 to 25 of 33

 
All times are UTC. The time now is 11:35.

Thu Feb 25 11:35:12 UTC 2021 up 84 days, 7:46, 0 users, load averages: 1.09, 1.36, 1.39

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2021, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.