YAFA=Yet another factorization algorithm :)
I don't know whether this is any good (I _think_ it's new) - comments welcome...
Suppose we wish to factor N.
Write N=x^(2^n)-1, s.t. x=(N+1)^(2^-n)
Then N=(x-1)(x+1)(x^2+1)(x^4+1)...(x^2^(n-1)+1)
Start w/ n=1, and evaluate all the 2^n combinations of products of brackets above, testing for exact division into N as you go.
After all unsuccessful 2^n attempts, increment n, and repeat.
Wednesday 12 March 2008
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