2021-08-26, 15:01 | #1 |
Jun 2021
110011_{2} Posts |
Inverse of FFT multiplication?
on the sample
101*109=11009 (1) let x=10 x^4+x^3+9 at the same time (x^2+1)*(x^2+9)=x^4+10*x^2+9 i.e. x^4+x^3+9-(x^3-10*x^2)=x^4+10*x^2+9 Bold polinomial exist for any composite. So if we find the bold for someone x, we can manage to factor number Seems still not good and complicated... Let x=100 (x+1)*(x+9)=x^2+10*x+9 and 11009=x^2+10*x+9 100 is lucky number? Yes and no. 100 = 101-1)) ok. Let x=98 11009=x^2+14*x+33=(x+3)*(x+11)=101*109 ...for some x, numbers start breaks to factors) in this simple form - take some x, make polynomial from given composite number N and try to factor, not numeric but in the algebraic - sometimes yield to results for factor N or N+1 or N-1)) P.S. I have no program for this part of idea. |
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