Is there anyone here that's fairly expert with the GMP libary on the Windows platform. I've recently started using the gmp.dll file in the distribution ...
7630
Phil Carmody
thefatphil
Aug 16, 2006 5:51 pm
Posted by: "kevinacres" research@... kevinacres ... Why can't you just use something like electric fence? Phil () ASCII ribbon campaign...
7631
Kevin
kevinacres
Aug 16, 2006 9:54 pm
... I just ran out of time in the day to further investigate the problem. However, the pitfall was my own. The usual 'cut and paste' trick. What I pasted...
7632
Robert
robert44444uk
Aug 17, 2006 9:42 am
... I was thinking about the fact it seems to take forever using the nextprime function. Although the range of P#/M+n doesnt ever get too large, before you hit...
7633
Ronny Edler
ronny_edler
Aug 17, 2006 2:44 pm
... Hi Kevin! None that I encountered... Since it seems that you have access to a compiler, simply download the latest source from http://www.swox.com/gmp/ and...
7634
Phil Carmody
thefatphil
Aug 18, 2006 3:09 am
Posted by: "Robert" rw.smith@... robert44444uk ... Cristiano recentl posted almost everything you need for a nextprime function to sci.crypt. I have...
7635
cino hilliard
hillcino368
Aug 26, 2006 9:16 am
I am trying to prove that all factors of numbers of the form 4k^2+1 are of the form 4m+1. Can someone help? Thanks, cino...
7636
Chris Caldwell
primemogul
Aug 26, 2006 12:15 pm
... If a prime p divides a^2+b^2 (but not b), then (a/b)^2 = -1 (mod p); so -1 is a quadratic residue and it follows that (-1)^((p-1)/2) = 1 (mod p) and hence...
7637
Chris Caldwell
primemogul
Aug 26, 2006 12:38 pm
... (or p = 2, which is not possible in your case.)...
7638
cino hilliard
hillcino368
Aug 26, 2006 9:36 pm
Thanks much Chris, I needed this to prove such things as n^3 + 7 != k^2 I will reference this posting where I use your proof. Cino...
7639
Phil Carmody
thefatphil
Sep 1, 2006 2:17 pm
As everyone knows, binomial coefficients are very smooth, which means that +/-1 terms have a reasonably high prime density. Looking at the top-5000, with a few...
7640
David Broadhurst
djbroadhurst
Sep 1, 2006 10:01 pm
... Cino is going to send us a proof that that there is no integer k such that k^2-7 is the cube of an integer. k^2-17 is the cube of an integer for the...
... Thanks, Phil! I couldn't do it on paper, because I don't know the ... [ (-2 : -3 : 1), (8 : 23 : 1), (43 : -282 : 1), (4 : 9 : 1), (2 : -5 : 1), (-1 : 4 :...
7643
David Broadhurst
djbroadhurst
Sep 2, 2006 7:33 am
... http://www.math.harvard.edu/~elkies/hall.html gives this wonderous result: 5853886516781223^3 - 447884928428402042307918^2 = 1641843 which is of a quality...
7644
Peter Moreton
pete_moreton
Sep 2, 2006 9:25 am
Hi, PFGW.EXE fails on my Windows 2003 X64 / AMD Opteron system, (unhandled win32 exception) - is this a known issue? So, I thought, no problem, I'll just...
7645
pete_moreton
Sep 2, 2006 9:26 am
Could someone explain to me why big Sophie Germain primes are described in the form "x * 2^P-1" - eg "2540041185*2^114729-1", when the definition of a...
7646
Kermit Rose
kermit1941
Sep 2, 2006 9:27 am
Message ... 1. Re: Factors of 4k^2 + 1 ...
7647
Norman Luhn
nluhn
Sep 2, 2006 1:50 pm
Hello ! P is here the number p=2540041185*2^114729-1. Not the exponent "114729". ... ___________________________________________________________ Telefonate...
7648
Pierre CAMI
pierrecami
Sep 2, 2006 3:52 pm
... described ... Sophie GERMAIN primes are primes P such that P is prime and 2*P - 1 is prime too Big primes are not easy to prove primes except when of the...
7649
Norman Luhn
nluhn
Sep 2, 2006 9:32 pm
Hello members! Is it true that: If p mod 4=3 and [2^p MOD (2*p^2+p)] - (2*p+1) = 0 then p and 2p+1 (possible!) Sophie Germain prime pair p,2p+1. We have 1 PRP...
7650
Norman Luhn
nluhn
Sep 2, 2006 9:36 pm
Hello members! Is it true that: * If p mod 4=3 and [2^p MOD (2*p^2+p)] - (2*p+[2]) = 0 then p and 2p+1 (possible!) Sophie Germain prime pair p,2p+1. We have 1...
... Very nice! In brief, No square is congruent to 7 modulo 8. (2*h+1)^3 + 8 = (2*h+3)*(4*h^2+3) != k^2 + 1 since no factor of k^2+1 is congruent to 3 modulo...
7653
Ignacio Larrosa Ca...
ilarrosa
Sep 3, 2006 9:39 am
Saturday, September 02, 2006 11:58 PM [GMT+1=CET], ... If a = 4m + 3 and a can be write as a = s^3 - t^2, there isn't solutions to that Mordell equation if s =...
7654
Paul Underwood
paulunderwooduk
Sep 6, 2006 8:03 pm
Hi, congrats to Larry Soule for the 139,672 digit Near-rep digit prime record: 99*10^139670-1: http://primes.utm.edu/primes/page.php?id=78434 Paul -- recent OU...
7655
Larry Soule
lsoulesbc
Sep 7, 2006 11:56 am
Thanks - I started searching at n=60k running on 4-5 machines starting about 3 week ago. I'm not sure how long it will last. I see 7 new NRD entries in the...
7656
David Broadhurst
djbroadhurst
Sep 9, 2006 9:31 am
I saw that a new Mersenne PRP is under test....
7657
Chris Caldwell
primemogul
Sep 9, 2006 12:59 pm
... Yes indeed. GIMPS checked the residue, redid the last part of the test, and all looked well. It is currently undergoing an independent verification on a...
7658
David Broadhurst
djbroadhurst
Sep 11, 2006 12:30 am
... I saw that there was speculation, in another place, that Curtis and Stephen might have had received a second bolt of lightning in Missouri. If that...
