144 III More on Divisibility
The proof of Proposition 12 illustrates howproblems involvingordinary integers
may be better understood by viewing them as integers in a larger field of algebraic
numbers.
We now return to the study of an arbitrary quadratic fieldQ(
√
d),wheredis a
square-free integer. For convenience of writing we putJ=Od. As in Chapter II, we
say thatε∈Jis aunitif there existsη∈Jsuch thatεη=1. For example, 1 and
−1 are units. The setUof all units is evidently an abelian group under multiplication.
Moreover, ifε∈U,thenalsoε′∈U.
Ifεis a unit, thenN(ε)=±1, sinceεη=1 impliesN(ε)N(η)=1. Conversely, if
ε∈JandN(ε)=±1, thenεis a unit, sinceN(ε)=εε′andε′∈J.(Note,however,
thatN(α)=±1 does not implyα∈J. For example, inQ(
√
− 1 ),α=( 3 + 4 i)/ 5 ∈/G,
althoughN(α)=1.) It follows that, whend≡2or3mod4,α=a+b
√
dis a unit if
and only ifa,b∈Zand
a^2 −db^2 =± 1.On the other hand, whend≡1 mod 4,α=a+b(
√
d− 1 )/2 is a unit if and only if
a,b∈Zand
(b− 2 a)^2 −db^2 =± 4.But ifb,c∈Zandc^2 −db^2 =±4, thenc^2 ≡b^2 mod 4 and hencec≡bmod 2.
Consequently, the units ofJare determined by the solutions of the Diophantine
equationsx^2 −dy^2 =±4orx^2 −dy^2 =±1, according asd≡1ord≡1 mod 4.
This makes it possible to determine all units, as we now show.
Proposition 13The units of O− 1 are ± 1 ,±i and the units ofO− 3 are ± 1 ,
(± 1 ±i
√
3 )/ 2. For every other square-free integer d< 0 , the only units ofOdare± 1.
For each square-free integer d> 0 , there exists a unitε 0 > 1 such that all units of
Odare given by±εn 0 (n∈Z).
Proof Suppose first thatd<0. Then only the Diophantine equations with the+signs
need to be considered. Ifd<−4, the only solutions ofx^2 −dy^2 =4arey=0,x=
±2. Ifd<−4orifd=−2, the only solutions ofx^2 −dy^2 =1arey=0,x=±1.
In these cases the only units are±1. (The groupUis a cyclic group of order 2, with
−1 as generator.) Ifd=−3, the only solutions ofx^2 −dy^2 =4arey=0,x=± 2
andy=±1,x =±1. Hence the units are±1,±ρ,±ρ^2 ,whereρ=(i
√
3 − 1 )/2.
(The groupUis a cyclic group of order 6, with−ρas generator.) Ifd=−1, the only
solutions ofx^2 +y^2 =1arey=0,x=±1andy=±1,x=0. Hence the units are
± 1 ,±i. (The groupUis a cyclic group of order 4, withias generator.)
Suppose next thatd>0. With the aid of continued fractions it will be shown in§ 4
of Chapter IV that the equationx^2 −dy^2 =1 always has a solution in positive integers
and, by doubling them, so also does the equationx^2 −dy^2 =4. Hence there always
exists a unitε>1. For any unitε>1wehaveε>±ε′,sinceε′=ε−^1 or−ε−^1 .If
ε=a+bω,whereωis defined as in Proposition 11 anda,b∈Z,thenε′=a−bωor
a−b−bω, according asd≡1ord≡1 mod 4. Sinceωis positive,ε>ε′yieldsb> 0
andε>−ε′then yieldsa>0. Thus every unitε>1hastheforma+bω,where