Number Theory: An Introduction to Mathematics

3 Periodic Continued Fractions 193

Proof Suppose first thatξ=[a 0 ,...,ah− 1 ] has a periodic continued fraction expan-
sion. Thena 0 =ah≥1 and henceξ>1. Furthermore, since

ξ=(ph− 1 ξh+ph− 2 )/(qh− 1 ξh+qh− 2 )

andξh=ξ,ξis an irrational root of the quadratic polynomial

f(t)=qh− 1 t^2 +(qh− 2 −ph− 1 )t−ph− 2.

Thusξis a quadratic irrational. Sincef( 0 )=−ph− 2 <0and

f(− 1 )=qh− 1 −qh− 2 +ph− 1 −ph− 2 > 0

(even forh=1), it follows that− 1 <ξ′<0. Thusξis reduced.
Ifξis a reduced quadratic irrational, then its complete quotientsξn, which are
all quadratic irrationals, are also reduced, by Lemma 0 withη =ξ′.Sinceξn′ =
an+ 1 /ξn′+ 1 and− 1 <ξn′<0, we have

an=− 1 /ξn′+ 1 .

Thusξn,ξn′are the roots of a uniquely determined polynomial

fn(t)=Ant^2 +Bnt+Cn,

whereAn,Bn,Cnare integers with greatest common divisor 1 andAn>0. Further-
more,D=Bn^2 − 4 AnCnis independent ofnand positive. Sinceξnis reduced, we have

ξn=(−Bn+

√

D)/ 2 An,ξn′=(−Bn−

√

D)/ 2 An,

where

0 <

√

D+Bn< 2 An<

√

D−Bn.

If we putg=

√

D,then−Bn∈{ 1 ,...,g}and, for a given value ofBn,thereareat
most−Bnpossible values forAn. Consequently the number of distinct pairsAn,Bn
does not exceed 1+···+g=g(g+ 1 )/2. Hence we must have

ξj=ξk,ξ′j=ξk′

for somej,ksuch that 0≤j<k≤g(g+ 1 )/2. Ifj=0, this already proves that the
continued fraction expansion ofξis periodic. Ifj>0, then

aj− 1 =− 1 /ξ′j=− 1 /ξk′=ak− 1

and hence

ξj− 1 =aj− 1 + 1 /ξj=ak− 1 + 1 /ξk=ξk− 1.

Repeating this argumentjtimes, we obtainξ 0 =ξk−j. Thusξhas a periodic continued
fraction expansion in any case.
If the period ish,sothatξ = [a 0 ,...,ah− 1 ], thenξ 0 ′ =ξh′ and the relation
an=− 1 /ξn′+ 1 implies that− 1 /ξ′=[ah− 1 ,...,a 0 ].