A simple method of prime factorization. Pull mod from mod!? In fact, I will explain what is correct by looking at the movement of numerical valuesby Nada 20230926

Hello. This is Nada from Seito Medical School. This time, we will talk about a powerful tool that performs prime factorization. Prime factorization? I've never heard of it. Look at the reality now. Get out of bed, brush your teeth, and go to school. yes. Look at reality. Let's start with something realistic. Prime factorization is a difficult problem to solve, similar to the one used in cryptography. This problem is related to a difficult mathematical problem called PvsNP, and current computers cannot solve the RSA encryption, which is a code for decomposition of prime factors. So, let's solve prime factorization using a new methodology. First of all, the tool we use is a mod. mod is also called remainder operation and is a tool related to remainders. 54=3mod17 indicates that the remainder when 54 is divided by 17 is 3. Furthermore, the problem 17×11=187 means that when a=11, there are 17 a's. Let's introduce a new way of thinking or a new set of values. mod3 means divide by 3. If you divide a multiple of 3 by 3, the remainder is 0. In other words, when 3a out of 11a is divided by 3, the remainder is 0. 17a=14a mod 3. If it is mod5, 5a=0mod5, 17a=12a mod5. What happens if we subtract these two equations? 11×17=1mod3, 11×17=2mod5. (1mod3)-(2mod5)=2a. In other words, 14a-12a=2a. Furthermore, arbitrarily, (2mod5)-(7mod9)=4a.

Here, subtract (2mod5)-(7mod9)=4a from both equations (1mod3)-(2mod5)=2a.

Then, (1mod3)+(7mod9) = -2a+(4mod5).

Here, 1mod3 means 11×17-11×3=154=1mod3, and 7mod9 means 11×17-11×9=88=7mod9, so the left side is 154+88=242. 4mod5 is 2mod5+2mod5, so 11×17 is multiplied twice. That is, (11×17-11×5)x2=264=4mod5. Looking at the total, it becomes 154+88=(-2)x11+242.

Now add (1mod3) and (7mod9). There are two 11x17s, so (11x17x2)mod(3+9)=2mod12=154+88=242. This is because we know that the number that can be calculated using this method is 11×17=187.

Then, (1mod3)+(7mod9)=(2mod12)= -2a+(4mod5).

This becomes (2mod12)-(4mod5)=242-264=-2a. (2mod12)-(11×17-4-5a)=-2a.

22-183mod12=7mod12=-7a, so (-7×7)a=(7×7)mod12. a=(-1)mod12=11mod12.

a=11, b=17, axb =187