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Old 2021-08-26, 15:01   #1
RomanM
 
Jun 2021

628 Posts
Default 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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