20200913, 18:03  #1 
Jun 2019
2×17 Posts 
factoring 2ⁿ2 equivalent to factoring 2ⁿ1(I think)
2^{n}1= a*b
factoring 2^{n}2 equivalent to factoring 2^{n}1 if this condition is met (a1) divided by (b1) then 2^{n}1 =1 mod b1 example M11 = 0 mod 23 and 88 =0 mod 22 2^{11}2 = 0 mod 22 M23 = 0 mod 47 2^{23}2 = 0 mod 46 M73 = 0 mod 439 2^{73}2 = 0 mod 438 M83 = 0 mod 167 2^{83}2 = 0 mod 166 M131 = 0 mod 263 2^{131}2 = 0 mod 262 Last fiddled with by baih on 20200913 at 18:42 
20200913, 18:25  #2 
Sep 2002
Database er0rr
111100010110_{2} Posts 

20200913, 18:42  #3 
Jun 2019
2×17 Posts 
JUST arror because 2304167 MOD 232 not equal to n1
SO sory 29 is false 2^{n}1= a*b this condition is IF (a1) divided by (b1) Last fiddled with by baih on 20200913 at 18:52 
20200914, 13:16  #4 
Feb 2017
Nowhere
1376_{16} Posts 
If p is prime then 2^{p}  2 is divisible by 2p. So if q divides 2^{p}  1, gcd(q1, 2^{p}  2) is divisible by 2p.
However, 2^{n}  2 is divisible by 2 but not by 4 if n > 1. If p is prime and the smallest prime factor q of 2^{p}  1 is congruent to 1 (mod 4), q1 does not divide 2^{p}  2. Consulting a table of factorizations of 2^{n}  1, I find that p = 29 (already noted) is the smallest p with this property. The next is p = 53 (q = 6361). 
20200917, 22:30  #5 
Jun 2019
2·17 Posts 
greatest common factor
its clear that for a composite numbre n = ab n1 divide gcf(a1,b1) if gcf are a big numbre we can use it for factoring n this what make the contruction of RSA numbre (beware) of making gcf small and does not help for factoring example n= 14111 = 103* 137 14110 = 0 MOD 34 and of course gcf(102,136) = 34 Last fiddled with by baih on 20200917 at 22:35 
20200917, 22:59  #6 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
3^{2}·11·101 Posts 
So
30526410117453380369827961 is a semiprime gcf(a1,b1) = 8 a free fact granted by me. Does this help you find a and b? n1 = 2^3×5×17×5827×1288671127×5978327543 so the gcf could be any combination of 1 or more of those. 
20200917, 23:06  #7 
Jun 2019
42_{8} Posts 
no
becaus 8 very smal i said earlier it needs to be very large to help 
20200918, 00:03  #8  
Mar 2019
2·3^{2}·11 Posts 
Quote:


20200918, 01:51  #9 
Jun 2019
2×17 Posts 
I DONT try to facto 2^12771 BY 2^12772
because i know the possibilty to find a large gcf(a1,b1) is very very small (excluded) approximately the gcf < 10 digts it wont do anything if the factor is large but this does not mean that the method is incorrect 
20200921, 07:11  #10  
"Jeppe"
Jan 2016
Denmark
2^{3}·3·7 Posts 
Quote:


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