gfactor(2) => 2; similarly for gfactor(5), etc.
Correct result is (1+%i)*(1-%i) etc.
Same thing for factor(2,q^2+1) -- but note that
factorization isn't necessarily unique in this case, e.g.
factor(6,q^2+5)=2*3=(1+q)*(1-q).
This appears to be a documentation bug (although it
would be nice to handle gaussian integer factorization).
Factorization of integers is special-cased in the code,
but there is no special case for factorization of gaussian
integers. There do exist simple algorithms for this.
Perhaps I'll implement one someday....
Maxima 5.9.0
Logged In: YES
user_id=501686
Need to change documentation for gfactor per comment above.