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

> -----BEGIN PGP SIGNATURE-----

> Version: GnuPG v1.2.4 (GNU/Linux)

>

> iD8DBQFAk8UoFXvAfvngkOIRApPlAJ9dmZZVwIuhJ1s0wZ0+ww3cUgmvjgCdGRtB

> JTgY0kJULWDT+ZlIcyb0w1E=

> =bk9N

> -----END PGP SIGNATURE-----

>

>

>

> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com

> The Prime Pages : http://www.primepages.org/

>

>

> Yahoo! Groups Links

>

>

>

>

>

>