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This question is part of on going research to factorize RSA 2048

The complete research context is convoluted and complex but I am trying to advance it by solving a small part of the research

consider this equation:

f1b = (2.Lr - 1) / (n - 2.f2b)

Important notes before the question:

  1. the dot is used as symbol for multiplication

  2. x is also used below for multiplication

  3. f1b = 367. So this is a known number.

  4. Lr is unknown but Lr mod f1b = 184 and since 184 x 2 = 368 then 2.Lr - 1 mod f1b = 0

  5. n mod f1b is 231. n is a known number.

  6. n is also the product of f2a and f2b which are both unknown. In the above equation we not dealing with f2a but you can see that f2b is in the denominator.

Is there enough information to know what the answer is for:

i) 2.f2b mod f1b, or

ii) f2b mod f1b?

Part of the context for this question can be read in an earlier post at the following math.stackexchange.com page:

Remainder, Short Invariance

At the end of this previous post I commented: Are there ways that I can pick my input factors and divisor r in such a configuration that will reduce the time (loop iterations)in searching for the unknown Lr .

In a live, large scale scenario:

a) n will be RSA 2048. So to know in advance what f2b mod f1b is ahead may be enough to factorize RSA 2048. Remember that I said above that n = f2a x f2b

b) f1b will be the first prime number below the square root of RSA 2048.

ziadcassim
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  • There might be a good question here, but it is incredibly hard to read. The choice of notation is responsible for much of this. For one, if $f1b$ is 367, then it doesn't need a complicated variable name: it can just be 367. – Randall Nov 03 '21 at 19:11
  • comment noted. The names are research names they do have meaning in the research context and I did provide some context by referencing the previous post. I also want to promote research into factorization of the RSA challenge numbers and I am not particularly attached to being the first to factorize RSA2048 if it can be done in second and minutes and not millions of years. By saying it in the research name I want to invite people to race to factorize this highest challenge number, if the math can support it. – ziadcassim Nov 04 '21 at 17:04
  • In the research context there is actually an f1a but it is not part of this equation.

    The input configuration used in the remainder, short invariance post had 4 numbers: 67 313 277 499

    After that post I changed the second number from 313 to 367. so then the input configuration is 67 367 277 499

    the names of the four numbers are: f1a f1b f2a f2b

    the square root of f2a x f2b is 371 and the first prime below 371 is 367. So 367 replaces 313. I want to choose a number that I know positively is above f2a. So then unless f2a is 367 it has to be below 367.

     – ziadcassim Nov 04 '21 at 17:54
  • In the present post I said that Lr mod 367 is 184. The Lr I am referring to is the new Lr after the swap. If I have knowledge of New Lr I can factorize RSA2048 in minutes But while I don't know what new Lr is I do know what Lr mod f1b is because New Lr mod f1b = (q+1) mod f1b and remember that q + 1 is the same before and after the swap of factors.

    Even though I don't what new Lr is I do know that the Lr Difference value (which is Start Lr - New Lr) mod f1b = 136. Lr difference is unknown. The 136 is known because Lr difference mod f1b = start Lr mod f1b

     – ziadcassim Nov 04 '21 at 18:03
  • Also, The previous input configuration in the remainder, short invariance post used r of 5764.

    Presently, in my attempt to generate new configuration properties that can be used to narrow the search for f2a or f2b I changed the r divisor to 134. 134 is 2 x 67 and the remainder, short invariance outputted is 67 which is also f1a.

    By fixing this way then after the swap the new Quotient from dividing dividend d by divisor q + 1 is (f2a-2).f1a. which is 275 x 67 = 18425.

     – ziadcassim Nov 04 '21 at 18:05

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