Number Theory: An Introduction to Mathematics

140 III More on Divisibility

Then

ρ(n,m)=ρ(r,m)=(r/m)=(n/m).

Sinceρ(m,n)=(m/n), this yields a contradiction. Thus,if n is an odd positive
integer,

G(m,n)=(m/n)G( 1 ,n)

for any odd positive integer m relatively prime to n.

In fact this relation holds also ifmis negative, since

G( 1 ,n)=(− 1 )(n−^1 )/^2 G( 1 ,n) and G(−m,n)=G(m,n).

(It may be shown that the relation holds also ifmis even.) As we have already obtained
an explicit formula forG( 1 ,n), we now have also an explicit evaluation ofG(m,n).

2 QuadraticFields

Letζbe a complex number which is not rational, but whose square is rational. Since
ζ/∈Q, a complex numberαhas at most one representation of the formα=r+sζ,
wherer,s ∈ Q.LetQ(ζ)denote the set of all complex numbersαwhich have a
representation of this form. ThenQ(ζ)is afield, since it is closed under subtraction
and multiplication and since, ifrandsare not both zero,

(r+sζ)−^1 =(r−sζ)/(r^2 −s^2 ζ^2 ).

EvidentlyQ(ζ) = Q(tζ)for any nonzero rational numbert.Conversely,if
Q(ζ)=Q(ζ∗),thenζ∗=tζfor some nonzero rational numbert.Forζ∗=r+sζ,
wherer,s∈Qands=0, and hence

r^2 =ζ∗^2 − 2 sζζ∗+s^2 ζ^2.

Thusζζ∗is rational, and so isζζ∗/ζ^2 =ζ∗/ζ.
It follows that without loss of generality we may assume thatζ^2 =dis a square-
free integer. Thendt^2 ∈Zfor somet ∈Qimpliest∈ Z.Ifζ∗^2 =d∗is also a
square-free integer, thenQ(ζ)=Q(ζ∗)if and only ifd=d∗andζ∗=±ζ.
Thequadratic fieldQ(

√

d)is said to berealifd>0andimaginaryifd<0. We
define theconjugateof an elementα=r+s

√

dof the quadratic fieldQ(

√

d)to be
the elementα′=r−s

√

d. It is easily verified that

(α+β)′=α′+β′,(αβ)′=α′β′.

Since the mapσ:α→α′is also bijective, it is anautomorphismof the fieldQ(

√

d).
Sinceα′=αif and only ifs=0, the rational fieldQis the fixed point set ofσ.Since
(α′)′=α, the automorphismσis an ‘involution’.
We define thenormof an elementα=r+s

√

dof the quadratic fieldQ(

√

d)to
be the rational number