Re: is it sufficient to solve factoring problem



However, if phi(N) = a* 2^i with some big
i, maybe the solution will cost much, right?

No, in fact, it will cost a little less.
Count the number of squares and of multiplications you have
to do for both the exponentiation X^K and the sequence X[i].

I initially mean if a<<2^i, then the probability of finding X^k = +/-
1 is small
, and then the finding process will redo many times.

.



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