- **Miscellaneous Math**
(*https://www.mersenneforum.org/forumdisplay.php?f=56*)

- - **factoring 2ⁿ-2 equivalent to factoring 2ⁿ-1(I think)**
(*https://www.mersenneforum.org/showthread.php?t=25958*)

factoring 2ⁿ-2 equivalent to factoring 2ⁿ-1(I think)2[SUP]n[/SUP]1= a*b
factoring 2[SUP]n[/SUP]-2 equivalent to factoring 2[SUP]n[/SUP]-1 if this condition is met (a-1) divided by (b-1) then 2[SUP]n[/SUP]-1 =1 mod b-1 example M11 = 0 mod 23 and 88 =0 mod 22 2[SUP]11[/SUP]-2 = 0 mod 22 M23 = 0 mod 47 2[SUP]23[/SUP]-2 = 0 mod 46 M73 = 0 mod 439 2[SUP]73[/SUP]-2 = 0 mod 438 M83 = 0 mod 167 2[SUP]83[/SUP]-2 = 0 mod 166 M131 = 0 mod 263 2[SUP]131[/SUP]-2 = 0 mod 262 |

[QUOTE=baih;556894]
8<-----SNIP------ M29 = 0 mod 233 2[SUP]29[/SUP]-2 = 0 mod 232 8<-----SNIP------ [/QUOTE] [CODE](2^29-2)%232 174 [/CODE] :ermm: |

JUST arror because 2304167 MOD 232 not equal to n1
SO sory 29 is false 2[SUP]n[/SUP]-1= [B]a*b[/B] [COLOR="Black"][B] this condition is IF (a-1) divided by (b-1)[/B][/COLOR] |

If p is prime then 2[sup]p[/sup] - 2 is divisible by 2p. So if q divides 2[sup]p[/sup] - 1, gcd(q-1, 2[sup]p[/sup] - 2) is divisible by 2p.
However, 2[sup]n[/sup] - 2 is divisible by 2 but not by 4 if n > 1. If p is prime and the smallest prime factor q of 2[sup]p[/sup] - 1 is congruent to 1 (mod 4), q-1 does not divide 2[sup]p[/sup] - 2. Consulting a table of factorizations of 2[sup]n[/sup] - 1, I find that p = 29 (already noted) is the smallest p with this property. The next is p = 53 (q = 6361). |

greatest common factor
its clear that for a composite numbre n = ab n-1 divide gcf(a-1,b-1) 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 |

So
30526410117453380369827961 is a semi-prime gcf(a-1,b-1) = 8 a free fact granted by me. Does this help you find a and b? n-1 = 2^3×5×17×5827×1288671127×5978327543 so the gcf could be any combination of 1 or more of those. |

no
becaus 8 very smal i said earlier it needs to be very large to help |

[QUOTE=baih;557258]no
becaus 8 very smal i said earlier it needs to be very large to help[/QUOTE] [url]http://factordb.com/index.php?query=2%5E1277-2[/url] is fully factored. Are you saying you can therefore factor 2^1277-1? |

I DONT try to facto 2^1277-1 BY 2^1277-2
because i know the possibilty to find a large gcf(a-1,b-1) 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 |

[QUOTE=mathwiz;557260][url]http://factordb.com/index.php?query=2%5E1277-2[/url] is fully factored. Are you saying you can therefore factor 2^1277-1?[/QUOTE]
I see an easy factorization of 2^n - 1 by induction: Suppose 2^(n-1) - 1 is factored. Then the factorization of 2^n - 2 is trivial. If we could somehow get the factorization from 2^n - 1 from that, ... /JeppeSN |

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