Thanks Senpai, will do.
Lads, can I get some help adding BigInteger into my Visual Studio? I've got the new code ready to run, but apparently don't have the following:
System.Windows.Forms
System.Numerics
BigInteger
Anyone willing to help me get up to speed can post in the thread below, and thanks in advance for helping this programming Newfag.
He means when you multiply (c * small prime) you get a new e value for each, with a new (e,1) to explore for factors. So based on your calcs we have 116 new (e,n) where c is a factor. Basically 116 iterations to a solution (I think) for RSA 2048.
Hello Anon, nice work!!! I've studied your method closely, and this is the verification method VQC posted code for here: >>8679
Nice work using it to verify our classics c145 and c6107. So we know the method of using c' = abq works.
Next we have this to work on: >>8779
The element you've verified in the new (e,n) is the BigN element.
Next we need to explore (e,1) to find the element with the same c'=abq or (an) or (prime b) or any other piece of info that can solve the problem.
For the c145 example, c'=5295 and e=49. (an) and (prime b) are the a and b values. I have a good idea for the method to do this, but have some commitments tonight. I'll be back later!
Let's generate (e,1) for the c'abq example for c6107 and see what we find in the first few elements.
Hello Anon! This post >>8772 is using the code VQC posted here >>8679.
The purpose is to make c bigger using (c * small primes). The code is used to verify that the new c' is valid.
We are using the Grid to create new c' values. When me multiply c with smaller primes, we get new (e,n) locations for each c'. When a=1, this is the BigN element. Our next step is to explore the new (e,1) values in each c' to see if the factors for c show up.
For c145, 145 * 5 = 725. New e=49. And the first element in (49,1) contains the solution. We get a=25 (which is (an)) and b=29 (prime b). So we shift to new locations in the grid each time we multiply c. Check out my screencap here: >>8800
c'=abq is the formula.
The paradigm shift for me personally is making c bigger by multiplying it with primes. RSA sized numbers are huge already, so I hadn't thought of that as a possible method. Now that I see it it makes perfect sense.
>For our friend 6107, only need first 3 primes in the series:
>[5, 11, 17]
c6107 * 5 = 30535, sqrt(30535) = 174 r259. So new e=259
Wut!? I accidentally did two square roots in a row on my calculator, and found something very interesting.
c6107 * 5 = 30535
sqrt(sqrt(30535)) = 13.219
So I pulled up e=13 in my Grid, and found 31 right there as a factor.
Making c bigger gives us more chances to find the factors.
Thanks Anon!
>>8811 (nice dubs Topol!)
IDK, I kinda like the VQC and his 12 disciples thing. Jesus is my Higher Power, so I get your point, but no need to get all grumpy Jan. Geez.
Hello AA! I think the main point of this c'=abq=qc is to get (prime a) or (prime b) to "fall out", not this smoothness BigN - n thing. For the small examples I've studied, we get elements in (e,1) where the solution is revealed. Multiplying c * q just gives new (e,n) to examine for factors. What are your thoughts on this?
I think we factorize each step in the series of primes that create q.
Meaning, check each (e,1) for each prime or combo of primes.
Up to q.
(An) or (prime a) or (prime b) will “fall out”.
It’s like tricking the Grid into giving us the answer, lol
The key is the series of primes leading up to q.
Hacking the Grid.
By making c bigger.
Fuck all this BigN - n shit, I want factors.