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On Saturday 01 May 2004 09:46, you wrote:

> Group...

>

> 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.

> 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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