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 2009-01-12, 00:34 #45 Mini-Geek Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 10AF16 Posts 2^(3*m)=(2^3)^m (2^3)^m=8^m 8^m=(7+1)^m Okay, following it so far... (7+1)^m==(0+1)^m mod 7 How is the 7 eliminated so we know that 8^m==1^m? 1^m=1 so 2^(3*m)==1 mod 7 And the rest makes sense, so it's just that one part I'm confused on.
2009-01-12, 03:10   #46
gd_barnes

May 2007
Kansas; USA

32×11×107 Posts

Quote:
 Originally Posted by Mini-Geek 2^(3*m)=(2^3)^m (2^3)^m=8^m 8^m=(7+1)^m Okay, following it so far... (7+1)^m==(0+1)^m mod 7 How is the 7 eliminated so we know that 8^m==1^m? 1^m=1 so 2^(3*m)==1 mod 7 And the rest makes sense, so it's just that one part I'm confused on.

Think of it like this:

(7+1) == (1 mod 7)
-and-
(0+1) == (1 mod 7)

therefore x represents the same integer 0<=x<=6 in both of the following:
(7+1)^m == (x mod 7)
-and-
(0+1)^m == (x mod 7)

therefore:
(7+1)^m == (0+1)^m mod 7

therefore:
8^m == 1^m mod 7

Last fiddled with by gd_barnes on 2009-01-12 at 03:11

2009-01-12, 06:22   #47
S485122

"Jacob"
Sep 2006
Brussels, Belgium

177710 Posts

Quote:
 Originally Posted by Mini-Geek (7+1)^m==(0+1)^m mod 7 How is the 7 eliminated so we know that 8^m==1^m?
1 == 8 == -6 == 15 == -13 mod 7 is the same as saying 1, 8, -6, 15 and -13 are multiples of 7 plus 1.

If we compute (7+1)^m = 7^m+m*7^(m-1)*1+ ... +m*7*1^(m-1)+1^m all terms except 1^m contain a power of 7 greater than 0. So we can say that (7+1)^m is a multiple of 7 plus 1.

More generally : (a*n+b)^m == b^m mod n. You could lookup some more explanation about modular arithmetics...

Jacob

 2009-01-13, 04:49 #48 Jens K Andersen     Feb 2006 Denmark 2·5·23 Posts From an unrelated PFGW search: 204912863*2^33333-2147 is composite: RES64: 69ED3420123456F2 (1.9872s+0.0003s) Not that much for a 2s test.
 2009-04-30, 23:36 #49 MooooMoo Apprentice Crank     Mar 2006 2×227 Posts I got 8 A's in one residue: 313*2^919519-1 is not prime. LLR Res64: B9AA9AE1AAAD1AAD Time : 1317.195 sec.
 2009-04-30, 23:43 #50 kar_bon     Mar 2006 Germany 1011011100102 Posts other 8er's: 1199*2^198096-1 is not prime. LLR Res64: CCBCCCF57770CCC0 1179*2^76893-1 is not prime. LLR Res64: 6C1D14411111CA41 1135*2^139165-1 is not prime. LLR Res64: C656B5A966666D76 1165*2^138973-1 is not prime. LLR Res64: F49A8666A66666F0 1179*2^170264-1 is not prime. LLR Res64: A8A2C562AAAAAA6E the last one with 6 A's in a row! Last fiddled with by kar_bon on 2009-04-30 at 23:45
 2009-05-01, 17:05 #51 MyDogBuster     May 2008 Wilmington, DE 22×23×31 Posts 3311*2^101894-1 CCCCCAC8484CF1AF 29.088 (Nice start) 10096*45^112584-1 93663508540e5a65 4912.17 (Almost all numeric) 10096*45^36114-1 df8a177777509fac 230.17 (Craps anyone)
2009-06-03, 17:57   #52
henryzz
Just call me Henry

"David"
Sep 2007
Liverpool (GMT/BST)

3·5·397 Posts

Quote:
 Originally Posted by henryzz a list up to n=1000 made using llr's predecessor prp Code: 4259877765*2^4+1 4259877765*2^10+1 4259877765*2^16+1 4259877765*2^139+1 4259877765*2^142+1 4259877765*2^328+1 4259877765*2^415+1 4259877765*2^583+1 4259877765*2^868+1 4259877765*2^883+1 these numbers are all prime: Code: (4259877765*2^4+1)/7 (4259877765*2^10+1)/7 (4259877765*2^16+1)/7 (4259877765*2^139+1)/7 (4259877765*2^142+1)/7 (4259877765*2^328+1)/7 (4259877765*2^415+1)/7 (4259877765*2^583+1)/7 (4259877765*2^868+1)/7 (4259877765*2^883+1)/7 edit: it works the other way as well i searched for primes of the form "(4259877765*2^n+1)/7" up to n=10k and then checked the residue for "(4259877765*2^n+1)" and it was always the same edit2: just realised that the numbers all have the same residue to each other in other prp bases as well though different to 3-prp
i just found this:
it explains why lots of numbers have the same residues

time to repeat my question:
does anyone know of a prp program that can display the full residue not just the RES64?
i could then use it to find prps hopefully
i expect the answer to be no but i would quite like a response

 2009-06-03, 21:28 #53 Cruelty     May 2005 22×11×37 Posts I've searched all my residues for the sequences of at least 6 digits / letters of the same kind: Code: 1515*2^618318-1 is not prime. LLR Res64: 5461D351111114AF --> 6 x 1 25*2^695867-1 is not prime. LLR Res64: 7D382AE72222229E --> 6 x 2 59*2^1625132-1 is not prime. LLR Res64: E3CAA3333333171E --> 7 x 3 736320585*2^715845-1 is not prime. LLR Res64: 9604AF6744444483 --> 6 x 4 2*3^184602-1 is not prime. RES64: 214E3DFFA875B519. OLD64: B59B17A74A777777 --> 6 x 7 736320585*2^93318-1 is not prime. LLR Res64: 18106436ABBBBBBA --> 6 x B
 2009-07-30, 23:49 #54 mdettweiler A Sunny Moo     Aug 2007 USA (GMT-5) 3·2,083 Posts Here's an interesting one that showed up today: 424*93^64337-1 is composite: RES64: [CAFFBBEEEA1063A4] (381.0392s+0.0157s)
 2009-09-21, 03:50 #55 mdettweiler A Sunny Moo     Aug 2007 USA (GMT-5) 624910 Posts Just now I was doing an off-the-wall search for MooMoo's "BEEF15BAD" residue with very small numbers (a fixed n search for k=3-1G, n=5, k*2^n+1), when I noticed a very weird pattern: Code: 168195*2^5+1 = 5382241 is prime! (trial divisions) 168221*2^5+1 = 5383073 is prime! (trial divisions) 168225*2^5+1 = 5383201 is prime! (trial divisions) 168231*2^5+1 = 5383393 is prime! (trial divisions) 168239*2^5+1 = 5383649 is prime! (trial divisions) 168245*2^5+1 = 5383841 is prime! (trial divisions) 168251*2^5+1 = 5384033 is prime! (trial divisions) 168269*2^5+1 = 5384609 is prime! (trial divisions) 168281*2^5+1 = 5384993 is prime! (trial divisions) 168293*2^5+1 = 5385377 is prime! (trial divisions) 168305*2^5+1 = 5385761 is prime! (trial divisions) This continues such that every number so far in my sieve file is prime. (I stopped the LLR testing at that point since I figured it was a pointless exercise, considering that it wouldn't be able to produce any composites and therefore useful residues.) I'm sure there's a simple mathematical explanation for what I'm seeing here. You know, though I hate to sound "crankish"...if there is a simple mathematical proof that all k*2^5+1 are prime, then this could lead to a very simple way to find a 100 million digit prime that would qualify for the EFF prize! Heck on spending 3+ years per number searching 100 million digit numbers through GIMPS when you can just find one this way. (Of course, I'm sure there's something I'm missing that would preclude this, otherwise someone would have won the prize by now.) Edit: I'm seeing this on k*2^7+1 as well. Last fiddled with by mdettweiler on 2009-09-21 at 03:56

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