- Apr 29, 2014
Hi i just little amateur in mathematics but working hard to find best methodology for factorzing semiprime.I believe every information is precious. Here i revealing some of my ideas. May be i miss little to express properly. Please give your valuable feedback.

Let N = pq be any semi-prime where p < q.

Let 'd' be the no. of digits of N. {For eg. if N = 21 then d = 2 since there are two digits 2 & 1.}

Let 'n' be any odd number.

Then Calculate the sequence

K = [ (2n^2 - 5n + 3)x(10^d) - (0x10^d + N) ] (mod p),

[ (2n^2 - 5n + 3)x(10^d) - (1x10^d + N) ] (mod p),

[ (2n^2 - 5n + 3)x(10^d) - (2x10^d + N) ] (mod p),

..........

.........[ (2n^2 - 5n + 3)x(10^d) - ((2n-4)x10^d + N) ] (mod p).

Then the sequence 'K' will be uniformly distributed either in order or disorder. {For eg. if p = 11, it is possible that K = 1,3,5,7,9,0 uniform distribution in order or else like K = 9, 5, 3, 7, 1, 0 uniform distribution not in order.}

This would have gave me a new methodology for factorizing N = pq where no.of digits of N is 'd'.

Choose any odd number 'n'

Then calculate the sequence

L = [ (2n^2 - 5n + 3)x(10^d) - (0x10^d + N) ],

[ (2n^2 - 5n + 3)x(10^d) - (1x10^d + N) ] ,

[ (2n^2 - 5n + 3)x(10^d) - (2x10^d + N) ] ,

..........

.........[ (2n^2 - 5n + 3)x(10^d) - ((2n-4)x10^d + N) ].

Let m = 0, 1, 2, 3, .....

Try to find GCD( (L +- m), N) which will give factor for N.

This methodology will be optimized for big integers by carefully choosing 'n' & 'm'