Re: integer n-roots like sqrtint()

[Bill Allombert <Bill.Allombert@math.u-bordeaux1.fr>]

On Mon, Feb 04, 2008 at 12:20:27PM +0100, Jeroen Demeyer wrote:
> Hello list,
> 
> I am wondering if there is an analogous function to sqrtint, but for 
> n-th roots?  With this I mean to compute floor(sqrtn(x,n)) exactly for 
> integers x and n.
> Unfortunately the obvious "floor(sqrtn(x,n))" does not work:
> 
> gp> \p100
>    realprecision = 105 significant digits (100 digits displayed)
> gp> floor(sqrtn(6^3,3))
> %33 = 5

Using round would be significantly more reliable...

If x is a perfect n-power, you can use ispower(x,n,&z);z

Cheers,
Bill.