20201229, 04:26  #1 
Dec 2017
F0_{16} Posts 
I (have not) found factors for 2^1099473911, I shot myself in the foot!
Although I really wanted to find a Mersenne Prime number with 2^1099473911, the facts are facts and it has factors other than 1 and itself. I spent a couple of weeks messing around with Pythonic code tweaking it to see if I could reveal factors. Well 2^1099473911 starts its factors low with the number 13 and produces a monster cofactor, the cofactor I cannot put in the spoiler, because its to large. I will also kindly leave the code which helped me find the factors. I think I somehow have broke cryptography to a certain degree by finding a factor and having prior knowledge for a giant number other than the kind folks here at GIMPS!
Here at this site you can see I'm running an LL on the the exponent 2^1099473911, but now I find it kind of fruitless. https://www.mersenne.org/report_expo...ll=1&ecmhist=1 You can use this site to divide or multiply numbers with high precision to see that I'm correct about finding the factor for 2^1099473911. http://www.javascripter.net/math/cal...calculator.htm Or if your good at python than I provide the factor for the most talked about exponent of 202021: Now if you wish to factor your own number quickly it can be done, however you need to download mprint and print the number you wish to factor with the code I will provide. Here is the site for mprint: http://www.apfloat.org/apfloat/ First here is the hidden factor: The code is next: 13 The code which found the factor WHICH beat GIMPS! Code:
#!/usr/local/bin/python # * coding: utf8 * import math import time import random import sys #y^2=x^3+ax+b mod n prime=[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271 ] # ax+by=gcd(a,b). This function returns [gcd(a,b),x,y]. Source Wikipedia def extended_gcd(a,b): x,y,lastx,lasty=0,1,1,0 while b!=0: q=a/b a,b=b,a%b x,lastx=(lastxq*x,x) y,lasty=(lastyq*y,y) if a<0: return (a,lastx,lasty) else: return (a,lastx,lasty) # pick first a point P=(u,v) with random nonzero coordinates u,v (mod N), then pick a random nonzero A (mod N), # then take B = u^2  v^3  Ax (mod N). # http://en.wikipedia.org/wiki/Lenstra_elliptic_curve_factorization def randomCurve(N): A,u,v=random.randrange(N),random.randrange(N),random.randrange(N) B=(v*vu*u*uA*u)%N return [(A,B,N),(u,v)] # Given the curve y^2 = x^3 + ax + b over the field K (whose characteristic we assume to be neither 2 nor 3), and points # P = (xP, yP) and Q = (xQ, yQ) on the curve, assume first that xP != xQ. Let the slope of the line s = (yP  yQ)/(xP  xQ); since K # is a field, s is welldefined. Then we can define R = P + Q = (xR,  yR) by # s=(xPxQ)/(yPyQ) Mod N # xR=s^2xPxQ Mod N # yR=yP+s(xRxP) Mod N # If xP = xQ, then there are two options: if yP = yQ, including the case where yP = yQ = 0, then the sum is defined as 0[Identity]. # thus, the inverse of each point on the curve is found by reflecting it across the xaxis. If yP = yQ != 0, then R = P + P = 2P = # (xR, yR) is given by # s=3xP^2+a/(2yP) Mod N # xR=s^22xP Mod N # yR=yP+s(xRxP) Mod N # http://en.wikipedia.org/wiki/Elliptic_curve#The_group_law''') def addPoint(E,p_1,p_2): if p_1=="Identity": return [p_2,1] if p_2=="Identity": return [p_1,1] a,b,n=E (x_1,y_1)=p_1 (x_2,y_2)=p_2 x_1%=n y_1%=n x_2%=n y_2%=n if x_1 != x_2 : d,u,v=extended_gcd(x_1x_2,n) s=((y_1y_2)*u)%n x_3=(s*sx_1x_2)%n y_3=(y_1s*(x_3x_1))%n else: if (y_1+y_2)%n==0:return ["Identity",1] else: d,u,v=extended_gcd(2*y_1,n) s=((3*x_1*x_1+a)*u)%n x_3=(s*s2*x_1)%n y_3=(y_1s*(x_3x_1))%n return [(x_3,y_3),d] # http://en.wikipedia.org/wiki/Elliptic_curve_point_multiplication # Q=0 [Identity element] # while m: # if (m is odd) Q+=P # P+=P # m/=2 # return Q') def mulPoint(E,P,m): Ret="Identity" d=1 while m!=0: if m%2!=0: Ret,d=addPoint(E,Ret,P) if d!=1 : return [Ret,d] # as soon as i got anything otherthan 1 return P,d=addPoint(E,P,P) if d!=1 : return [Ret,d] m>>=1 return [Ret,d] def ellipticFactor(N,m,times=5): for i in range(times): E,P=randomCurve(N); Q,d=mulPoint(E,P,m) if d!=1 : return d return N first=True # ENTER YOUR NUMBER TO FACTOR FOR n: n = 2047 # n = if (len(sys.argv)>1): n=int(sys.argv[1]) # print (n,'=',) for p in prime: while n%p==0: if (first==True): print (p,) break first=False else: print ('x',p,) break n/=p m=int(math.factorial(2000)) while n!=1: k=ellipticFactor(n,m) n//=k if (first==True): print (k,) first=False # else: print ('x',k,) break Last fiddled with by ONeil on 20201229 at 04:55 
20201229, 04:43  #2  
Aug 2006
3^{2}·5·7·19 Posts 
Quote:
Code:
> Mod(2,13)^1099473911 %1 = Mod(10, 13) 

20201229, 04:47  #3 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}·1,531 Posts 

20201229, 05:03  #4  
Dec 2017
11110000_{2} Posts 
Quote:
Why when I multiply 13 after I divide it by 13 the result = 2^1099473911 so what you are saying is I should go the full distance on the Lucas Lemer test and their is still hope. This is strange I have not stopped the processing the number yet. 

20201229, 05:09  #5 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}×1,531 Posts 
Since you love Python so much, try this:
Code:
print((2 ** 109947391  1) % 13) Code:
print((2 ** 11  1) % 23) Last fiddled with by retina on 20201229 at 05:10 
20201229, 05:48  #6  
Dec 2017
2^{4}·3·5 Posts 
Quote:
I have proof and I'm posting the video. First off you need to use mprint to print out the entire number then like this: n = hugeexponent % 13 print(n) n=0 video will be posted ok! 

20201229, 05:53  #7 
"Curtis"
Feb 2005
Riverside, CA
1282_{16} Posts 
You're so bad at math you don't even understand why you're wrong when told you're wrong.
No Mersenneform number 2^p1 can have a factor smaller than p. It's not possible. This is easily proved, easily enough that you ought to be able to understand the proof if you go read it. In fact, no factor smaller than 2p+1 is possible. That's why it is obvious even to noncoders that your code is garbage. Your ignorance and trolling are tiring. 
20201229, 06:01  #8  
Dec 2017
360_{8} Posts 
Quote:


20201229, 06:09  #9 
Dec 2017
2^{4}×3×5 Posts 
Here is the video proof that when you mod the exponent by 13 its zero. period:

20201229, 06:22  #10 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
9495_{10} Posts 
I went to the site that you used.
http://javascripter.net/math/calcula...calculator.htm I entered 2 in the X field and 109947391 in the Y field, then hit the X^Y button and got: This computation would take too long as the answer. Most likely your input was truncated. Try again. 
20201229, 06:25  #11  
Dec 2017
360_{8} Posts 
Quote:


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