The problem came from this:
I'd like to have one unique routine that simplifies both of the following expressions to zero.
sqrt((1+sqrt(2))^2)-(1+sqrt(2));
sqrt((1+sqrt(2))^2)-sqrt(1+2*sqrt(2)+2);
I tried
sqrt((1+sqrt(2))^2)-(1+sqrt(2)); --> output: 0
but
sqrt((1+sqrt(2))^2)-sqrt(1+2*sqrt(2)+2); --> output -sqrt(2^(3/2)+3)+sqrt(2)+1
Compared with this:
sqrt(expand((1+sqrt(2))^2))-(1+sqrt(2)); --> output: sqrt(2^(3/2)+3)-sqrt(2)-1
and
sqrt(expand((1+sqrt(2))^2))-sqrt(1+2*sqrt(2)+2); --> output: 0
Thanks for your help!
Am 23.11.2012 um 22:33 schrieb Stavros Macrakis <macrakis at alum.mit.edu>:
> I believe that the only factorization that Maxima does over quadratic integers is over Z[sqrt(-1)], the Gaussian integers, using the function gcfactor. In general (as you probably know), quadratic integers don't have unique factorizations.
>
> If you're interested in this topic, you might want to look at: http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/quadraticgrad.pdf . It would be great if you could implement the relevant algorithms for Maxima!
>
> -s
>
> On Fri, Nov 23, 2012 at 3:39 PM, Ren? K?lin <renekaelin at gmx.ch> wrote:
> Hello
>
> Is it possible to factor 1+2*sqrt(2)+2? I'd like to get (1+sqrt(2))^2.
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