Re: Simple integer factorization algorithm



a^2 = (119)(10^2)^{-1} - 1 mod 11 = 8 mod 11,

119/100 -1 mod 11 = 1.19 - 1 mod 11 != 8 mod 11

I don't even understand your example...
though I doubt it will help to understand your Idea.. if you are willing to do Maths it would be good to learn how to express yourself a bit clearer.. Thats an important part of science in general.. A paper or any scientifical result is worthless if the inventer can't express himself. Ununderstandable papers simply get rejected.

If I understood correctly the flaw in your algorithm lies in that decrementing of k for checks...
you actually need only polynomial ammount of decrements respective to the number. But that is not enought ... you need logarithmic so your algorithms does not become exponential. (Number representations are logarithmic in their size.)
But may be I just misunderstood you .. its all possible.
You were not that clear...
besides you are offtopic.

Christian
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