## 14814Re: [PrimeNumbers] the start of AKS-algorithm?

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• May 1, 2004
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On Sat, 1 May 2004, Décio Luiz Gazzoni Filho wrote:
> On Saturday 01 May 2004 09:46, you wrote:
> > If I wanted to prove that some number N is not equal to the form a^b, then
> > wouldn't I have to start by finding the square root, the third root, the
> > fifth root, and every odd-number root ... and not stop until the result was
> > either integral or less than 2?
>
> Sure, that's a possibility. Not the fastest either, but since you're using
> AKS, you don't care the least about speed after all.

I do believe that the proposed method is the fastest known (except that
you only need to check roots of prime order of course). But I am willing
to see arguments otherwise. See "Detecting perfect powers in essentially
linear time" by Daniel Bernstein at http://cr.yp.to/papers/powers-ams.pdf

>
> > What would be the maximal order of this type of calculation if N is 'm'
> > digits long?
>
> Let's call this largest power k. Then we know that N^(1/k) < 2, and we also
> know that N ~ 10^m. Thus (10^m)^(1/k) < 2, or 10^(m/k) < 2. Now take
> logarithms to base 10 on both sides, you get m/k < 0.3. Thus m < 0.3k, or k >
> m/0.3, or finally, k > 3.3m.
>
> Décio
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