Re: A factoring algorithm

hart_wb@yahoo.com wrote:
> Hi,
>
> I have a new factoring algorithm which seems to be very fast with
> certain kinds of numbers.
>
> For example it will factor numbers which are the product of two primes
> which are both of the "same kind", and fairly close together (say are
> the same length to within 3 or 4 digits -though that isn't an absolute
> requirement if you do some fiddling and have a little more time), where
> by "same kind" I mean that they are both of the form:
>
> a*b^n+c where a, b and c are "small"

Your post lacks specifics. Define "small"

Is n fixed? You said in another post that b is fixed.

If, in fact, a and c are bounded by a polylog function of b^n Then
a factorization of N = (a1 x + c1)(a2 x + c2) where a_i, c_i <
log(x)^k for
some k, then this factorization is readily found in polynomial time
(where
x = b^n.)

Numbers of this form are also susceptible to the Special Number Field
Sieve.

It is *easy* to construct limited sub-sets of the integers, possessing
special
form whose factorizations are easy.

The world does not need another O(N^1/4) algorithm.
Received on Tue Jan 17 16:49:13 2006