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Factoring Mersenne and other numbers using tri-math

  1. May 2, 2005 #1
    Let T(n)=n(n+1)/2. As discovered in my earlier efforts an expanded
    representation of the values for A-F that solve the formula T(A*n+B)=
    (C*n+D)*(E*n+F), for all integer n is as follows:
    A= 4t^2*m^2 + 6t^2*m + 2t^2
    B= (2t^2-2t)*m^2 + (3t^2-3t-2)*m +(t^2-t-2)
    C= 2t^2*m^2 + 4t^2*m + 2t^2
    D= (t^2-t)*m^2 + (2t^2-2t-1)*m + (t^2-t-1)
    E= 4t^2*m^2 + 4t^2*m + t^2
    F= (2t^2-t)*m^2 + (2t^2-2t-2)*m + (t^2-t-2)/2

    I noted that n can be any real number(I suspect that n could even be
    complex!!), for instance if m=2,t=2 and n=1/3 the equation T(A*n+B)=
    (C*n+D)*(E*n+F) becomes

    T({8/3}*{8+6+1}+16+8) = ({8/3}*{4+4+1}+8+6+1)*({4/3}*{16+8+1}+16+4)
    =T(64)=39*(4/3*25 + 20) = 13*(100+60) = 65*32 = 64*(64+1)/2

    Notice how the factors 13 and 5 of 65 conveniently become separated
    between the two factors of T(64). This method could also work if we let
    A*n +B be one less than any other number that needs factoring.

    As an example of factoring 63 which is a Mersenne number, we can try
    letting A*n+B = -64 by setting n = -11/15. Then my tri-equation becomes

    T(-64)=-189/5 * -160/3 = 63*64/2 but this didn't factor the 63.

    However, let T=2,M=3,n=1/2 then my tri-equation becomes

    T(-64)= 36*56 which sucessfully factors the 63 as one merely has to divide each factor by the factors 4,8 of 64/2.

    I believe this method has a lot of potential. Any comments
     
  2. jcsd
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