Experimenting with this:
%i11) factor(x^8+a+1,a^4+a+1);
2 2 4 2
(%o11) (x - a) (x + a) (x + a )
(%i12) factor(x^8+a+1,a^8+a+1);
Maxima encountered a Lisp error:
Error in MACSYMA-TOP-LEVEL [or a callee]: Caught fatal error [memory may be
damaged]
Automatically continuing.
To reenable the Lisp debugger set *debugger-hook* to nil.
(%i13) factor(x^7+a+1,a^7+a+1);
6 5 2 4 3 3 4 2 5 6
(%o13) (x - a) (x + a x + a x + a x + a x + a x + a )
(%i14)
I hope the powers show up correctly.
RRogers
-----Original Message-----
From: Stavros Macrakis [mailto:macrakis at gmail]
Sent: Monday, October 31, 2005 10:33 AM
To: Lodder, Josje
Cc: maxima@math.utexas.edu
Subject: Re: [Maxima] factor(x^8+1)
You can factor x^8+1 by adjoining a root of unity, e.g.
factor(x^8+1,a^2+1) gives a product of two quartics
factor(x^8+1,a^4+1) gives a product of four quadratics
In both cases, 1/... is integrable. You can then substitute back in the
value of a.
But the result is rather messy.