20111126, 09:55  #1 
Nov 2011
2^{2}·3 Posts 
Some Properties of Mersenne Number Factors
a) b) Am I correct ? 
20111126, 16:15  #2 
Dec 2010
Monticello
5×359 Posts 
Our resident curmudgeon is out.
Can you define the(wedge) operator so more of the amateurs can follow a little better? 
20111126, 16:47  #3 
Nov 2011
2^{2}×3 Posts 
definition of logical conjunction which math symbol is :

20111126, 19:36  #4 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
I'm not sure can't say the gcd stuff for sure but since; so the is true. for the gcd to be correct you have to prove if then . you may complete the thought process if you want I've been sitting or walking down at the market for 6+ hour after 8 hours of poor if any sleep.

20111126, 20:33  #5 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
a) Prime factors q of Mp with p an odd prime are always 1 (mod p) and +1 (mod 8), so k==0 (mod p) in your notation is correct. If 3(k1), then 8k == 2 (mod 3) and q = 8k+1 == 0 (mod 3), which would make q not a prime factor. So gcd(k1, 3)=1.
b) q = 8k1 = lp+1, so 8k = lp+2, so 8k == 2 (mod p) <=> 4k == 1 (mod p). If k+1 were divisible by 3, etc., as above, just with signs changed. Last fiddled with by akruppa on 20111126 at 20:33 
20111126, 21:56  #6 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Code:
b=[];forstep(p=6*80+5,1,[4,2],if(isprime(p),forstep(y=1,100000*p+1,if(p%6==1,[4*p,2*p],[2*p,4*p]),if(isprime(y)&&(2^p1)%y==0,if(isprime((y1)/(2*p)),b=concat(b,[[(y1)/(2*p),p]]);break()))))) 
20111127, 05:33  #7  
Nov 2011
2^{2}×3 Posts 
Quote:


20111127, 16:16  #8 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
could be I tried using mod 24 before but I can't remember where on here or otherwise I put the result. I started from the fact that
Last fiddled with by science_man_88 on 20111127 at 16:18 
20111128, 08:09  #9 
Nov 2011
2^{2}·3 Posts 

20111128, 09:08  #10 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10011011101001_{2} Posts 
You don't need FT to prove that. Simple algebraic calculus shows it. All factors of mersenne numbers M=2^p1 with p prime (in fact odd p is enough) can only be of the form 2*k*p+1 for some k, and they can only be of the form 8*x+1 or 8*x1 for some x. So, you split in two cases:
(1). The factor is 8*x+1 and 2*k*p+1, equate them, take the 1 out, simplify by 2, so kp=4x, and p is prime and 4 is a prime power, it follows that x contains p and k contains 4, so k is a multiple of 4, and all factors of this form are in fact (by renotating k): 8*k*p+1, for any k bigger or equal to 1. (2). The factor is 8*x1 and 2*k*p+1, equate them, move the 1 on the other side, simplify by 2, etc etc, you will reach some conclusion like the factors can only be (renotating k): 8*k*p+s*p+1, where k bigger or equal to 0, and s is the double of the complement of p (mod 4). That is: if p=1 (mod 4) then s=6 and if p=3 (mod 4) then s=2. So this case splits in: (2a). Factor is 8*k*p+2*p+1 if p=3 (mod 4) and (2b). Factor is 8*k*p+6*p+1 if p=1 (mod 4), where k>=0 in both situations. You can call the factors of the form (1) "plus factors", and the factors of the form (2a) or (2b) "minus factors". This is all you can make from the discussion in this thread, and from all discussions related to the form of the factors. There are no other relations you can deduct from it for the form of the factors, even if you take them mod 4, mod 3, mod 24, mod whatever, unless you discover something really NEW about the form of the factors. All attempts lead more or less to this, after more or less circling around the tail. This is the strongest (known) conclusion that you can get, and includes all the stuff with the powers of 3 or other things written before in this thread. There were plenty of discussion on the forum about it. This is just wasting time. edit: Please remark that it is not necessary that a mersenne M to have "plus factors" (it can have none), but if M is composite, it will always have an odd number of "minus factors", that is: it will have at least one minus factor (or 3, or 5, etc). The "minus factors" can not be in even number, otherwise multiplying them will lead to a number which is 1 (mod 8) and M is 7 (mod 8)  impossible. That is because multiplying by "plus" factors does not change the class (the number stays 1 mod 8 if it was 1 mod 8, and it stays 7 mod 8 if it was 7 mod 8). So only multiplying with "minus" factors can change the class. This means "it could make sense" to search for "that minus factor", and some searching was indeed conducted to look for it, as you can sieve them 4 times higher, but the disadvantage is that nobody tells you that the "minus" factor is the lowest (generally, they are not, ex: for M29, "plus" factors are 233 and 2089, and the "minus" factor is 1103). Last fiddled with by LaurV on 20111128 at 09:55 
20111128, 13:47  #11  
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
Quote:
Code:
(09:43)>forprime(x=5,100,print((2^x2)/(2*x))) 3 9 93 315 3855 13797 182361 9256395 34636833 1857283155 26817356775 102280151421 1497207322929 84973577874915 4885260612740877 18900352534538475 1101298153654301589 16628050996019877513 64689951820132126215 3825714619033636628817 58261485282632731311141 3477359660913989536233495 816785180559426160758185055 (09:43)>((2*k*p+1)*(2*j*p+1))/(2*x) %1 = (4*j*k*p^2 + (2*k + 2*j)*p + 1)/(2*x) (09:44)>((2*k*p+1)*(2*j*p+1))/(2*p) %2 = (4*j*k*p^2 + (2*k + 2*j)*p + 1)/(2*p) (09:44)>(2*k*p+1)*(2*j*p+1) %3 = 4*j*k*p^2 + (2*k + 2*j)*p + 1 (09:45)>((2*k*p+1)*(2*j*p+1)1)/(2*p) %4 = 2*j*k*p + (k + j) Last fiddled with by science_man_88 on 20111128 at 13:51 

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