From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: 17th roots of 1 Date: 1 Jul 1998 20:37:59 GMT In article , Lee Davidson wrote: >Let R+ = sqrt(17+4sqrt(17)), R- = sqrt(17-4sqrt(17)). Then this >expression occurs in the step from quartics to quadratics: > >sqrt( 1/4 * ( R+- +- 1 )( R+- +- sqrt(17) ) ) > >+- signs depend on the roots you're looking at. > >I tried looking for some way to write the expression within the outer >sqrt as a perfect square, but couldn't find any. Is this the simplest >way of writing this? It's pretty simple already, actually, considering. You want to know whether the numbers ( 1/4 * ( R+- +- 1 )( R+- +- sqrt(17) ) ) are perfect squares -- but in what ring did you mean this? Note that ((R+-)^2 - 17)/(+-4) = sqrt(17) so that sqrt(17) already lies in the rings Q[ R+ ] and Q[ R- ]. Moreover, (R+)(R-) = sqrt(17^2 - 4^2*17) = sqrt(17), so that each of these rings contain the generator of the other, that is Q[ R+ ] = Q[ R- ]. Now you are asking whether certain elements of this ring are squares. I take it you mean, are they the squares of other elements in this ring. Well, if x = y^2 then norm(x)= norm(y)^2, but if I've calculated correctly, these elements x have norm 17, not a square. So x is not a square in this ring. (You can calculate the norm in stages: it's norm = norm1 o norm2 where these new functions are the norms to and from the intermediate field Q[sqrt(17)]. Since each is a quadratic extension, each norm(X) is computed by multiplying X * X', where X' is obtained from X by changing the sign of the generator R+ or sqrt(17), respectively. In maple: ( 1/4 * ( Ra + ep1 )*( Rb + ep2* sqrt(17) ) ); "*subs({Ra=-Ra,Rb=-Rb},"); expand("); simplify(",{ep1^2=1,ep2^2=1}); 2 2 1/16 (Ra - 1) (Rb - 17) simplify(",{Ra^2=17+4*ep1*sqrt(17),Rb^2=17-4*ep2*sqrt(17)}); 1/2 -17 ep1 ep2 - 4 ep2 17 #(That's now norm2(x) lying in the intermediate field. Now repeat.) "*subs(sqrt(17)=-sqrt(17),"); expand("); simplify(",{ep1^2=1,ep2^2=1}); 17 dave