Hen e, for xed m, it was natural to ask fora ompletelist ofnegative fundamental dis riminants d su h that h( d

Imaginary quadrati elds

with small odd lass number

by

Steven Arno, M. L. Robinson and

Ferrell S. Wheeler (Bowie, Md.)

1. Introdu tion. Let dbethedis riminantofanimaginaryquadrati

eld with lass number h( d). As is well known, Gauss [9℄ onje tured

that h( d) tends to innitywith d. Hen e, for xed m, it was natural to

ask fora ompletelist ofnegative fundamental dis riminants d su h that

h( d) = m. This problem is usually referred to as Gauss' lass number

problemorGauss' lassnumberm problem.

In 1934,Heilbronn[14℄ su eeded inprovingGauss' onje ture, thereby

pla ing the lass number problem on rm ground. The following year,

Siegel[20℄showedthatforany">0thereexistsa onstant

"

>0su hthat

h( d) >

"

d 1=2 "

as d ! 1. Unfortunately, neither result was ee tive,

and no further progress was made until the 1950's when Heegner [13℄ of-

feredasolutionforthe lassnumber1problembasedonnewideasfromthe

theoryof modularfun tions. It isinteresting to re allthat Heegner'sproof

was generally dis ounted until the \gaps" in his argument were explained

manyyears later(see [5℄,[8℄,[21℄,[24℄). Intheinterimperiod,however, the

rst a epted proof of the lass number1 problemwasgiven byStark [23℄

in 1966. Shortlythereafter, Baker [2, 4℄ foundanother proof based on the

theory of trans endental numbers. In 1971, Baker [3, 4℄ and Stark[26{28℄

independentlyresolvedthe lass number2 problemaswell. However, there

seemed to be littlehopeof generalizingthese methodsto solvehigher lass

numberproblems.

In 1976, Goldfeld [10℄ presented a deep and entirely unexpe ted result

whi h provided the framework for a general atta k on the lass number

problem. He showed that if there exists a Weil urve whose asso iated

L-fun tionhasazeroofatleastthethirdorderat s=1,thenforany">0

1991 Mathemati sSubje tClassi ation: Primary11R;Se ondary11Y,11J.

Key words and phrases: binary quadrati forms, imaginary quadrati elds, lass

numbers,dis riminants.

there existsan ee tively omputable onstant

"

su h that

(1:1) h( d)>

"

(logd) 1 "

:

Of ourse, theutilityofGoldfeld'sresult dependedon ndingan appropri-

ate ellipti urve. Andthough one would expe t to nd a Weil urve with

a high order zero at s=1 based on the elebrated onje ture of Bir h and

Swinnerton-Dyer [5℄, te hni al diÆ ulties kept things on hold for several

years. Finally, in 1983 Gross and Zagier [11℄ were able to show that er-

tain urves must have a zero of at least the third order at s = 1, thereby

ompleting theatta k of Goldfeld.

Goldfeld's proof was later simplied by Oesterle [19℄, who provided,

among other things, expli it onstants for Goldfeld's theorem. As a re-

sult,Oesterlewasableto ompletethe lassnumber3problemaswell. The

lass number 4 problem was solved by the rst author [1℄ who ombined

new te hniques with the well-known methods of Stark [22℄, Montgomery{

Weinberger [18℄, and Oesterle [19℄. In hindsight, [1℄ ontains a prototype

for thepartitioning of minima(of redu edquadrati formsof dis riminant

d), whi hplaysa ru ialrole inthispaper.

The entral on ern of this paper is the lass number m problem for

small, odd m. The aforementioned partitioning of minima enables us to

signi antly improve on earlier estimates. Our results are summarized in

thefollowingtheorem.

Theorem 1. For ea h odd integer m satisfying 5  m  23, the lass

number m problem is solved. For ea h su h m, a omplete list of nega-

tive fundamental dis riminants d for whi h h( d) = m an be found in

Appendix A.

Let d be the dis riminant of an imaginary quadrati eld with lass

number h( d). In Table 1 we present the number of elds satisfying

h( d)=m and thelargest su h d forea h odd m satisfying1m23.

Table 1. Upperboundondsatisfyingh( d)=m

m #ofd max. d

1 9 163

3 16 907

5 25 2683

7 31 5923

9 34 10627

11 41 15667

13 37 20563

15 68 34483

17 45 37123

19 47 38707

21 85 61483

23 68 90787

The paper is organized as follows. In x 2 we use Oesterle's [19℄ expli it

onstants to produ e a d

3

(m) su h that if d  d

3

(m), then h( d) 6= m.

Inx 3 weprovidethe theoreti aljusti ation forthepartitioningof minima

mentioned above. In x 4 we prove several te hni al lemmas on erning an

auxiliaryfun tionwhi hareneeded inx5.

We divide x5 into three subse tions, ea h of whi h rules out a ertain

range of fundamental dis riminants. In x 5.1, following the approa h of

Montgomery{Weinberger [18℄, we produ e a d

2

(m) su h that if d

2 (m) 

d  d

3

(m), then h( d) 6= m. At the end of x5.1 we note that the meth-

ods of x5.1, even when pushed to their limits, do not redu e the range

of admissible dis riminants d to the point where a omputationally inten-

sive sieve (like the one introdu edinx6) an be usedto ompletethe lass

number m problem when m > 7. This shows that some new argument is

ne essary. In x5.2 we exploit the partitioning of minima introdu ed in x3

to produ e a d

1

(m) su h that if d

1

(m)  d  d

2

(m), then h( d) 6= m.

When m is odd and 5  m  13, d

1

(m) is small enough to allow the

lass numberm problemto be ompleted with the omputationally inten-

sive sieve of x 6. In x5.3 we use a more sophisti ated version of the ar-

guments in x5.2 to produ e a d

0

(m) su h that if d

0

(m)  d  d

1 (m),

then h( d) 6= m. When m is odd and 15  m  23, d

0

(m) is small

enough to allow the lass number m problem to be ompleted with the

omputationallyintensivesieve ofx 6. We ombinethe partitioningof min-

ima with the aforementioned sieve to omplete the proof of Theorem 1

inx6.

Notation. Let Z, Z +

, Q, R and C denote the ring of integers, the set

of positiveintegers,theeld ofrationalnumbers,theeld ofreal numbers,

andtheeldof omplexnumbers,respe tively. Letpandqdenoteprimesin

Z +

. TheKrone kersymbolisdenotedbyeither m

n

or(mjn),dependingon

whi h ismore onvenient. Asis ustomary, we let!(n) denote thenumber

ofdistin tprimedivisorsofnandd(n)thetotalnumberofpositivedivisors

of n. Finally,let e(x)=e 2ix

.

The number d denotesanegative fundamentaldis riminantor, equiv-

alently, thedis riminant of an imaginaryquadrati numbereld. In other

words, we have either d  1 (mod 4) and d is square-free, or 4jd and

d=4 2 or 3 (mod4) and d=4 is square-free. 

1

denotes the real primi-

tive hara terwith 

1

(n)=( djn).

A binary quadrati form

Q(x;y)=ax 2

+bxy+ y 2

of dis riminant d=b 2

4a isredu ed ifitsatiseseither

Note thatthisimplies

(1:3) a(d=3)

1=2

:

Let

Q

d

=fQ(x;y)=ax 2

+bxy+ y 2

:b 2

4a = d; Qisredu edg

denotethenitesetofredu edbinaryquadrati formswithdis riminant d.

Noteh( d)=jQ

d

j. Thenotation P

Qd

denotesasumoverallQ2Q

d along

withthe asso iated oeÆ ientsa, b,and .

ThemodiedBesselfun tionofthese ondkindoforder zeroisgivenby

(1:4) K

0 (z)=

1

\

1 e

(z=2)(t+t 1

) dt

t

(<z>0):

2. Thehigh range. Asmentionedinx1,Oesterle[19℄providedexpli it

onstantsforthe

"

inGoldfeld'sinequality(1.1). Indeed,if disanegative

fundamental dis riminant with lassnumberh( d) and

#(d)= Y

pjd





1 b2

p

p

p+1



;

wheretheprodu tistakenoverall primedivisorsp ofd withtheex eption

of thelargest primedivisor,then

#(d)logdCh( d)

whereC =55 if(d;5077)=1and C =7000 otherwise.

Inordertoevaluate#(d)whenh( d)isodd,re allthattheFundamental

Theoremof Generadueto Gauss [9℄impliesthat

2

!(d) 1

jh( d):

Ifh( d)isodd,thenthepre edingresultimpliesthatd hasonlyoneprime

divisor. Sin e d is a dis riminant, we see that d is either 4, 8, or

p forsome odd primep3 (mod 4). It followsthat when h( d)is odd,

#(d)=1. Sin eh( 4)=h( 8)=1,wehave

(2:1) h( d) isoddand h( d)>1)d is primeand d3 (mod 4) :

Hen e, when h( d) is odd and h( d) >1, (2.1) impliesthat (d;5077)=1

(for5077 isa primeequalto 1 modulo 4),and we have

logd55h( d):

Thus,form2f5;7;:::;23g, wehave h( d)>m ifdd

3

(m)whered

3 (m)

Table2. h( d)6=mfordd

3 (m)

m d

3 (m)

5 10 120

7 10 168

9 10 215

11 10 263

13 10 311

15 10 359

17 10 407

19 10 454

21 10 502

23 10 550

Note that when h( d) is even, d may be omposite. This allows the

possibility that a minimum a of a redu ed form ould satisfy (a;d) > 1.

This introdu es many te hni al diÆ ulties, the least of whi h is a smaller

valueof#(d). Forthese reasons,we onneourattentionto the asewhere

h( d) isodd.

3. Minima results. The oeÆ ient a is referred to as the minimum

of theformax 2

+bxy+ y 2

2Q

d

,whilethemultisetofminimafora given

negative fundamentaldis riminant d isdenoted by

M

d

=fa:ax 2

+bxy+ y 2

2Q

d g:

From(1.2)itiseasytoseethat1o ursinM

d

pre iselyon e. Wehen eforth

refer to 1 2 M

d

as the prin ipal minimum sin e 1 is the minimum of the

prin ipalform(i.e.,eitherx 2

+(d=4)y 2

orx 2

+xy+((d+1)=4)y 2

dependingon

theparity ofd). Furthermore,we notethat allelementsof M

d

arepositive

by(1.2).

Lemma 1. If a2M

d

,then any positive divisorof ais also in M

d .

Proof. Leta

1

>0beadivisorofa. Ifa

1

=athereisnothingtoprove.

Hen e,we may assumea

1

a=2. It followsfrom (1.3) that

(3:1) a

1

(d=12) 1=2

:

Letax 2

+bxy+ y 2

2Q

d

sothatb 2

4a = d. Itfollows aton ethat the

quadrati ongruen e z 2

 d (mod 4a

1

) is solvable. Hen e, there exists

a b

1

su h that b 2

1

 d (mod4a

1

); with 2a

1

< b

1

 2a

1

. Further, this

range an besharpenedto ensurethat a

1

<b

1

a

1

byrepla ingb

1 with

b

1 2a

1 if a

1

< b

1

 2a

1 and b

1

with b

1 +2a

1

if 2a

1

< b

1

 a

1 . Let

1

=(b 2

1

+d)=(4a

1

) and observe that

1

d=(4a

1 )> a

1

using (3.1). From

(1.2) we see thatthequadrati form Q=a x 2

+b xy+ y 2

isredu ed.

Lemma 2. Let a satisfy 2  a  (d=4) 1=2

and g d(a;d) = 1. Then

a2M

d

if and onlyif every prime divisor p of asatises ( djp)=1.

Proof. First, assumethat a2M

d

. Sin e b 2

4a = d, we seethat

(3:2) b

2

 d (mod4a):

If2ja, then we seeat on ethat b 2

 d (mod 8). We knowthat d must

be odd be ause a iseven and, by hypothesis, g d(a;d) =1. Sin e 1 is the

onlyodd square modulo 8, we have d 1 (mod 8). Thus( dj2) =1. If

p is any odd prime dividing a, it follows at on e from (3.2) that b 2

 d

(modp). Thus( djp)=1,and the\onlyif"dire tion isproved.

Conversely, assume that ( djp) =1 forea h prime divisor p of a. If p

is odd, then the ongruen e z 2

 d (mod p) is solvable, and byHensel's

lemma, the ongruen e z 2

 d (modp 

) is solvable for all  2 Z +

. If

p=2,notethat( dj2)=1impliesd 1 (mod 8). Hen e,the ongruen e

z 2

 d (mod 8) is solvable. Furthermore, it iswellknown that solutions

withz1 (mod 8) an beliftedto asolutionofthe ongruen e z 2

 d

(mod2 

)forall 2Z +

. Hen e,bytheChineseRemainderTheorem,there

existsabsu hthatthe ongruen eb 2

 d (mod 4a)issolvable. Reasoning

asin theproof of Lemma1, there is infa t su h ab with a<ba. Let

= (b 2

+d)=(4a). If b 6= 0, then > d=(4a)  a sin e a  (d=4) 1=2

by

hypothesis. Ifb=0,then =d=(4a)a. Inevery ase, ax 2

+bxy+ y 2

is

redu ed, anda2M

d .

Lemma 3. Suppose h( d) is odd. If a>1 and ax 2

+bxy+ y 2

2Q

d ,

then ax 2

bxy+ y 2

isa distin t member of Q

d .

Proof. Sin e a>1anda2M

d

,we knowthath( d)>1. Hen e,(2.1)

impliesthat d is prime. Sin e ax 2

+bxy+ y 2

is a redu edform, we know

that ax 2

bxy+ y 2

will be a distin t redu ed form unless (i) b = 0, (ii)

b=a, or(iii)a= . If(i) is true,then4a =d, ontradi ting thefa tthat

d is prime. If(ii)is true,then a 2

4a =a(a 4 ) = d. Sin ea>1 and

d is prime,we have a=d,whi his impossiblesin ea<(d=3) 1=2

. If(iii)is

true,thenb 2

4a 2

=(b 2a)(b+2a)= d. Sin eb0in ase(iii)anddis

prime,we knowthat2a b=a+(a b)=1. This leadsto a ontradi tion

sin ea>1byhypothesis andabby(1.3).

Lemma 4. Suppose h( d) is odd. Let a 2 Z +

be odd and satisfy

a(d=4) 1=2

. If a2M

d

,then aappears in M

d

exa tly 2

!(a)

times.

Proof. If a = 1, it is easy to see from (1.2) that a appears pre isely

on e in M

d

. Thus, we may hen eforth assume that a> 1. Of ourse, this

impliesthat h( d) >1 aswell. We want to ount the number of integers

b su h that ax 2

+bxy+((b 2

+d)=(4a))y 2

2 Q . By (1.2) this is just the

numberofintegersbthatsatisfy(b 2

+d)=(4a) 2Zwitheither a<ba<

(b 2

+d)=(4a) or0ba=(b 2

+d)=(4a). Assume, for themoment, that

(b 2

+d)=(4a)2Z. Sin e h( d)>1, we seefrom (2.1) that d3 (mod4).

It followsat on e thatb isnonzero. Using thehypothesis,a(d=4) 1=2

,we

then dedu e ad=(4a) <(b 2

+d)=(4a). It follows that we want to ount

thenumberofintegersbthat satisfyb 2

 d (mod 4a)with a<ba.

From(2.1),weknowthatdisprime. Sin ea(d=4) 1=2

<d,weseethat

g d(a;d)= 1. Let p be any prime divisor of a. By Lemma 2, we see that

( djp)=1. Sin e ais odd, we knowthat p is also. Hen e, the ongruen e

z 2

 d (mod p)hasexa tly two solutions. By Hensel's Lemma, we know

that the ongruen e z 2

 d (mod p 

) has exa tly two solutions for all

 2 Z +

. Also, sin e d  1 (mod 4), we know that the ongruen e z 2



d (mod 4) has exa tly two solutions. Thus by the Chinese Remainder

Theorem,weseethattherearepre isely2

!(a)+1

integersbthatsatisfyboth

(3:3) b

2

 d (mod4a)

and 2a<b2a. Noweitherbothbandb+2asatisfy(3.3) orneither do.

Hen e,thenumberof bthatsatisfy(3.3)with 2a<b ais equalto the

numberof b that satisfy (3.3) with 0 <b  a. A similarargument with b

and b 2a shows that the number of b that satisfy (3.3) with a< b 2a

is equal to the number of b that satisfy (3.3) with a <b  0. It follows

thatexa tly 2

!(a)

integersb satisfy(3.3) with a<ba,thereby proving

Lemma4.

Lemma 5. If a>1,a2M

d

and (a;d)=1,then a>(d=4) 1=h( d)

.

Proof. Suppose pja and p is an odd prime. Then b 2

 d (mod p)

implies ( djp) = 1. Also, if 2ja, then b 2

 d (mod 8). Hen e, d  1

(mod8), so that ( dj2) = 1. Thus, if p is any prime dividing a we have

( djp)=1. ThisimpliesthatpsplitsinQ(

p

d),hpi=}

1 }

2

with}

1 6=}

2 .

Thus,} h( d)

1

=hi is aprin ipal ideal,and  62Z. It followsthat p h( d)

=

N(}

h( d)

1

) = N(hi) d=4. Sin e a p, it follows that a (d=4) 1=h( d)

.

Note that a >1 impliesh( d)> 1, sothat (d=4) 1=h( d)

62Z. The lemma

follows.

It follows from (1.3),(2.1) and Lemma 5that

(3:4)



d

4



1=h( d)

<a



d

3



1=2

(a2M

d

nf1g; h( d)>1 is odd):

To improve on these bounds, we separate the minima to a ertain extent

usingthemultipli ativestru tureof M

d

developed inLemmas 1{4.

Assumethath( d)>1andh( d)isodd. FromLemma3weknowthat

every a2M nf1g appears an even numberof times inM . Also, by (3.4)

ifd>2

h( d)+2

,then2is nota minimum. Itthenfollows byLemma 1that

a must be odd if a 2 M

d

. In the remainder of this se tion assume that

d>2

h( d)+2

. DenethemultisetM



d by

(3:5) M



d

=fa:ax 2

+bxy+ y 2

2Q

d

;a>1;b>0g:

From Lemma3 we have

jM



d

j=(h( d) 1)=2:

Definition 1. Apartition ford isalistof(h( d) 1)=2pairsoffun -

tions(l

a (d);u

a

(d)),ea h in reasingind, orrespondingto the(h( d) 1)=2

elementsinM



d .

Definition 2. We saya partitionford oversM

d if

l

a

(d)au

a (d)

forevery a2M



d .

We begin with a simple example. Suppose that h( d) = 5. Let p be

thesmallestnonprin ipalmemberof M

d

. By Lemma1,p isprime. Inpast

investigationsofminimaonegenerallyusedtheresult ontainedinLemma5,

whi h statesthat all minimaa, ex epta=1,mustsatisfy a>(d=4) 1=5

. In

parti ular,theminima ouldsimultaneouslybe small,ea h minimumlying

loseto thebound(d=4) 1=5

. However,usingLemma2andLemma4,wesee

thatifp isabout(d=4) 1=5

insize,thenthere areexa tlytwo redu edforms

withminimumpandtworedu edformswithminimump 2

andone redu ed

form with minimum 1, whi h provides us with ve redu ed forms. This

ex ludes the possibility that any other minimum is simultaneously small.

Usingsimilar reasoningit isnot hard to see that ifh( d) =5,then M

d is

overed byone of thefollowingthree partitions:

1 (d=4) 1=5

p(d=4) 1=4

(d=4) 2=5

p 2

(d=4) 1=2

2 (d=4) 1=4

p(d=4) 1=3

(d=4) 1=4

a(d=3) 1=2

3 (d=4) 1=3

a(d=3) 1=2

(d=4) 1=3

a(d=3) 1=2

For ea h xed value of m=5;7;:::;23, a set of partitions overing all

possible M

d

is given in Appendix B. In these tables, p and q denote the

rstand se ondsmallestprimeminimainM

d

,respe tively,whileadenotes

a generi member of M

d

, v= d=4 and w= d=3. In order to simplify our

presentation of a overing partition,we now introdu e some additional no-

tation. Thenotation(3p.),forexample,denotesthattheinequalitiesforthe

rstthree powersoftheprimearetriviallyinferred. Similarly,thenotation

notation, the overingpartitions forM

d

when h( d)=5 aregiven by

1 v

1=5

pv 1=4

(2p:)

2 v

1=4

pv 1=3

v 1=4

aw 1=2

3 v

1=3

aw 1=2

(2)

In order to givea better ideaof howthepartitions inAppendixBwere

generated, we will go throughthe details of partition number 10 for lass

number23(Table B10). The partitionisgiven by

10 v 1=10

pv 1=8

(4p:) v 3=16

qv 1=4

(2p:) v 23=80

pqv 3=8

(2)

v 31=80

p 2

q v 1=2

(2) v 3=8

aw 1=2

Assume the smallest nonprin ipal minimum p satises v 1=10

 p  v 1=8

;

then p 2

;p 3

;p 4

are all  (d=4) 1=2

, implying that p;p 2

;p 3

;p 4

2 M

d

, whi h

a ounts for 8 minima. If the se ond smallest minimum q satises q 

(d=4) 3=16

, then q;q 2

;pq;pq 2

;p 2

q are  (d=4) 1=2

, implying that they are in

M

d

. Butthisa ountsfor16newminima,whi hwouldmake25total. Thus,

v 3=16

q. Assume q (d=4) 1=4

. Now, q;q 2

;pq;p 2

q (d=4) 1=2

,so we have

a ountedfor21minima. Ifanyfurtherminimaasatisfya(d=4) 3=8

,then

pa(d=4) 1=2

wouldimplythatthere aremorethan23 minima. Therefore,

theremaining two minimamust satisfy(d=4) 3=8

a(d=3) 1=2

. Allof the

partitions in the tables of Appendix Bare omputed in this same fashion.

Byusingadisjointset ofassumptionsontherst andse ond largestprime

minimapandq,thesesets ofpartitions anbeseento overallpossibleM

d .

4. Propertiesof anauxiliaryfun tion. Inthisse tionweprovetwo

te hni al lemmas on erningtheauxiliaryfun tionF dened by

(4:1) F(x)=

p

x 1

X

n=1 d(n)K

0

(nx) (x>0);

whereK

0

(z) is denedin(1.4).

On several o asions inx5.2 and x 5.3 we willneedto omputea urate

approximations to F at ertain small arguments. For su h purposes, the

following rude generalizationof [18, Lemma7℄) suÆ es.

Lemma 6. If x>0,N isa nonnegative integer,and

N

(x)=F(x) p

x N

X

n=1 d(n)K

0 (nx);

then

j

N (x)j

2

p

x

(1+log(N +1+2=x))e

(N+1)x=2

Proof. Sin e x>0, wesee from (4.1) and (1.4) that

0x 1=2

N (x)

1

X

n=N+1 d(n)

1

\

1 e

nxt=2 dt

t

= 1

X

n=N+1 d(n)

1

\

nx=2 e

u du

u :

Usingpartialsummation,we have

x 1=2

N (x)

1

\

(N+1)x=2 e

u



X

n2u=x d(n)



du

u :

Fory1we have

X

ny

d(n)= X

ny



y

n



y X

ny 1

n

y(1+logy);

sothat

jx 1=2

N (x)j

1

\

(N+1)x=2 e

u 2u

x

(1+log (2u=x)) du

u :

An integration bypartsgives

jx 1=2

N (x)j

2

x



(1+log(N+1))e

(N+1)x=2

+ 1

\

(N+1)x=2 e

u du

u



:

Note thatthefun tiong, denedby

g(v)= 1

\

v e

u du

u e

v

log(1+1=v);

isin reasing forv>0,and lim

v!1

g(v)=0. Hen e,

1

\

(N+1)x=2 e

u du

u

e

(N+1)x=2

log



1+ 2

(N +1)x



;

and thelemma follows.

The next lemma willplayan importantrole inx 5.2 and x5.3.

Lemma 7. F(x) is a stri tly de reasing fun tionof x for x>0.

Proof. From(4.1) we have

F(x)= p

x 1

X

n=1 d(n)K

0

(nx)= p

x 1

X

n=1 X

djn K

0

(nx)= p

x 1

X

d=1 1

X

m=1 K

0 (dmx);

where K is the modied Bessel fun tion of the se ond kind of order zero

given by (1.4). Hen e,we get

F(x)= p

x 1

\

1



1

X

d=1 1

X

m=1 exp



dmx

2

(t+t 1

)



dt

t

= 1

\

1



1

X

d=1

p

x

exp d(t+t

1

)

2 x

1



dt

t :

Forb>0,dene

f

b (x)=

p

x

e bx

1 :

Sin e x 7! x=(e xb

1) de reases on (0;1), we know that x 7! f

b

(x) also

de reaseson (0;1). Lemma 7follows immediately.

5. The medium range. Let  be a real primitive hara ter modulo

k for some integer k > 1. In the ase where k is a negative fundamental

dis riminantwe take (n)=( kjn)(see [7,p. 40℄). Dene

A

d (s)=

X

Q

d

(a)a s

and

P

k (s)=

Y

pjk (1 p

s

):

In1966Stark[22℄exploitedaformulaforthezetafun tionofaquadrati

numbereld(i.e.,(s)L(s;

1

))toshowthatifatenthfundamentaldis rim-

inant d of lass number 1 existed, then d > exp(2:2
10 7

). Later Stark

[25℄ developed a formula for L(s;)L(s;

1

) analogous to the formula for

(s)L(s;

1

). Montgomery and Weinberger [18℄ exploited this formula to

obtainsimilarresultsfor lassnumbers2 and 3. Indeed,if(k;d)=1,then

(5:1)



k p

d

2



s 1=2

(s)L(s;)L(s;

1 )=T

d

(s)+T

d

(1 s)+U

d (s)

where

T

d (s)=



k p

d

2



s 1=2

(s)(2s)P

k (2s)A

d (s);

(5:2) U

d (s)=

4 p



k X

Q

d a

1=2 1

X

n=1 K

s 1=2



n p

d

ak



n s 1=2

V

Q (s;n);

and

V

Q

(s;n)= X

y 1 2s

<



k

X

j=1

(Q(j;y))e



jn

ky



e



bn

2ak



:

Let s

0

= 1=2+it

0

, with t

0

> 0, be a zero of L(s;). Substitutings =s

0

into (5.1) gives

T

d (s

0 )+T

d (s

0

)= U

d (s

0 ):

ApplyingtheS hwarzRe e tion Prin ipleto T

d

implies

2jT

d (s

0

)j os (argT

d (s

0

))= U

d (s

0 );

whi h,inturn,gives

(5:3) jsinarg(iT

d (s

0 ))j=







 U

d (s

0 )

2T

d (s

0 )







 :

The method forthe middlerange onsists inshowing thatthisequality

isfalseforlargeintervalsofdundertheassumptionthath( d)issomexed

oddinteger. Indeed,forxed k and t

0

,denethe onstants



1

=t

0

=2;

(5:4)



2

=t

0 log



k

2



+arg fi (1=2+it

0

)(1+2it

0 )P

k

(1+2it

0 )g;

(5:5)



3

=2j (1=2+it

0

)(1+2it

0 )P

k

(1+2it

0 )j:

(5:6)

To show (5.3) isfalse, allweneed to showis that

(5:7) jsin(

1

logd+

2

+argA

d (s

0 ))j>

jU

d (s

0 )j



3 jA

d (s

0 )j

:

5.1. Therange d

2

(m)d d

3 (m)

Lemma 8. Let t2R +

. Suppose m=h( d) is odd and xed. Then

(5:8) jA

d

(1=2+it)j1

m 1

(d=4) 1=(2m)

:

Furthermore, if d>maxf4e 2m

;4(m 1) 2m

g,then

(5:9) jargA

d

(1=2+it)j

t(1 1=m)log(d=4)

(d=4) 1=(2m)

(m 1) :

Proof. Both (5.8) and (5.9) are trivially true if m = 1, so assume

m>1. Usingthelowerboundin(3.4) gives

jA

d

(1=2+it) 1j= 



 X

Q

d

;a6=1

(a)a 1=2 it







 X

Q

d

;a6=1 a

1=2

(5:10)



m 1

(d=4) 1=(2m)

: (5:11)

For the remainder of the proof assume d > maxf4e 2m

;4(m 1) 2m

g.

Sin e d>4(m 1) 2m

,we knowfrom (5.11) that

jA

d

(1=2+it) 1j<1;

so that A

d

(1=2+it) lies in the right half plane. Hen e, when we write

argA

d

(1=2+it)in(5.9)andbelow,we an,withoutlossofgenerality,assume

we are dealing with the prin ipal value of the argument. Let L be the

line segment joining 1=2 to 1=2+it. Then an equation for L is given by

`(u)= 1=2+iu, 0 u t. Sin e d >4(m 1) 2m

, we knowby (5.8) that

A

d

does notvanishon L. Furthermore,A

d

(s) isan entirefun tion of s. It

follows [17, p. 218℄ that

\

L A

0

d (z)

A

d (z)

dz= t

\

0 A

0

d

(1=2+iu)

A

d

(1=2+iu)

idu=logA

d

(1=2+iu)j t

0 :

Evaluatingthe right-handsideand takingimaginarypartsyields

jargA

d

(1=2+it)j= 







= t

\

0 A

0

d

(1=2+iu)

A

d

(1=2+iu) idu







 (5:12)

t max

0ut 





 A

0

d

(1=2+iu)

A

d

(1=2+iu) 





 :

Note that

jA 0

d

(1=2+it)j= 



 X

Q

d

;a6=1

(a)a 1=2 it

loga 



 (5:13)

 X

Q

d

;a6=1 a

1=2

loga:

Now, x 1=2

logx is a de reasing fun tionof x for x >e 2

. Using (3.4) and

thehypothesis d>4e 2m

,itthen followsfrom (5.13) that

(5:14) jA

0

d

(1=2+it)j

(1 1=m)log(d=4)

(d=4) 1=(2m)

:

Using(5.14) and (5.8) in(5.12) yields(5.9), and thelemma isproved.

Lemma 9. Let t2R. If k >1 is an oddsquare-freeinteger, then

(5:15) jU

d

(1=2+it)j 4

d 1=4

X

rjk 3

!(r)

2

!(k) !(r)

r 1=2

X

a2Md F



 p

dr 2

ak



:

Proof. Fromthedenitionof U

d

in(5.2) wededu e at on ethat

(5:16) jU

d

(1=2+it)j 4

p



k X

Q a

1=2 1

X

n=1 K

0



 p

dn

ak



jV

Q

(1=2+it;n)j:

Usingan argument ofWeil[30, his inequality(5)℄,Montgomery and Wein-

berger [18, Lemma7℄have shownthat

jV

Q

(1=2+it;n)j2

!(k)

k 1=2

X

yjn

Y

pj(y;n=y;k) p

1=2

2

=2

!(k)

k 1=2

X

rjk 2

!(r)

r 1=2

X

yjn

(y;n=y;k)=r 1;

be ausek issquare-free. Sin e (y;n=y;k)=r impliesr 2

jn, we have

jV

Q

(1=2+it;n)j2

!(k)

k 1=2

X

rjk

r 2

jn 2

!(r)

r 1=2

d(n):

Insertingthisinto (5.16) gives

(5:17) jU

d

(1=2+it)j

 4

p



k 1=2

X

a2M

d 1

a 1=2

X

rjk 2

!(k) !(r)

r 1=2

1

X

n=1

r 2

jn d(n)K

0



 p

dn

ak



:

Letn=r 2

m andnotethatd(n)d(r 2

)d(m)=3

!(r)

d(m) sin er issquare-

free. From (5.17) we have

jU

d

(1=2+it)j

 4

p



k 1=2

X

a2Md 1

a 1=2

X

rjk



3

2



!(r)

2

!(k)

r 1=2

1

X

m=1

d(m)K

0



 p

dr 2

m

ak



:

Applyingdenition(4.1) to theinnersumgives theresult.

Corollary. Let t2R andm =h( d). If k>1 is an odd square-free

integer,then

jU

d

(1=2+it)j 8k

1=2

logk

3 1=4

 1=2

d 1=4

(m 1+e

(d 1=2

p

3)=(2k)

) Y

pjk (2+3p

3=2

):

Proof. Repla eF inLemma9withtheupperboundgiveninLemma6

withN =0 to get

jU

d

(1=2+it)j 8k

1=2

 1=2

d 1=2

X

rjk



3

2



!(r)

2

!(k)

r 3=2

 X

a2M a

1=2



1+log



1+ 2ak

d 1=2

r 2



e

d 1=2

r 2

=(2ak)

:

In the inner sum, the a = 1 term is treated separately. When a > 1 we

use theinequalitya(d=3) 1=2

. In both ases, we also usethe inequalities

r1,d3,and



1+log



1+ 2k

 p

3



e

 p

3=(2k)

logk (k2)

to nishtheproof ofthe orollary.

Fortheremainderofx5.1,weassumethereexistsadis riminant dwith

thefollowingproperties:

(I) h( d)=m,where m2f5;7;9;:::;23g is xed;

(II) d

2

(m)d10 850

,where d

2

(m)isgiven inTable3 neartheendof

thissubse tion.

Our goal is to show that (5.7) is true for d with a suitable hoi e of k

and s

0

. For thispurpose, we needa smallzero, s

0

= 1=2+it

0

,of L(s;).

Weinberger [31℄ has omputed several su h zeros, ea h orresponding to a

dierent value ofk. In thisse tion,we use

k =115147 and t

0

=0:003157614

where the absolute error in t

0

is less than 10 8

, but we only make use of

therst 4 signi antpla es. From (5.4){(5.6) we thenhave



1

=0:001579; 

2

=0:02875; 

3

=555:8;

wherethese approximationsarea urate to thenumberof pla esshown.

Sin e m23 and d d

2

(m)10 63

byTable 3,the right-hand sideof

(5.8) is positive. Hen e, letting t=t

0

inLemma 8, we see from (5.8) that

jA

d (s

0

)j doesnot vanish. Thus, using (5.8) and the orollary to Lemma 9

witht=t

0

,wehave

jU

d (s

0 )j



3 jA

d (s

0 )j

R

2 (d);

where

R

2 (d)=

8k 1=2

logk(m 1+e

(d 1=2

p

3)=(2k)

) Q

pjk

(2+3p 3=2

)



3 3

1=4

 1=2

d 1=4

(1 (m 1)=(d=4) 1=(2m)

)

:

ClearlyR

2

(d)is de reasingind sothat

(5:18)

jU

d (s

0 )j



3 jA

d (s

0 )j

R

2 (d

2 (m))

sin edd

2

(m). UpperboundsforR

2 (d

2

(m)) aregiven inTable3.

Sin e 5  m  23 we have maxf4e 2m

;4(m 1) 2m

g = 4(m 1) 2m

<

10 63

<d

2

(m). Hen e,all ofthehypotheses ofLemma 8hold fort=t

0 and

m2f5;7;:::;23g. We dedu e from(5.9) that

jargA (s )j (d);

where



2 (d)=

t(1 1=m)log(d=4)

(d=4) 1=(2m)

(m 1) :

It iseasy to see that

2

(d) isde reasingford4e,sowe ertainlyhave

(5:19) jargA

d (s

0 )j

2 (d

2 (m))

sin edd

2

(m). Upperboundsfor

2 (d

2

(m)) aregiven inTable3.

Let 

2

(m) bedened by



2

(m)=

1 logd

2

(m)+

2 

2 (d

2 (m))

and

2

(m)bedened by

2

(m)=

1 log10

850

+

2 +

2 (d

2 (m)):

Lower bounds for

2

(m) and upperboundsfor

2

(m) aregiven inTable 3.

UsingTable3,thefa t thatd

2

(m)d,and (5.19), we have

0<

2

(m)

1

logd+

2

+argA

d (s

0 ):

Usingthefa tthat d10 850

,(5.19), and Table 3,we have



1

logd+

2

+argA

d (s

0 )

2

(m)<:

Hen e,

(5:20) jsin(

1

logd+

2

+argA

d (s

0

))jminfjsin

2

(m)j;jsin

2 (m)jg:

Lowerboundsforjsin

2

(m)j andjsin

2

(m)j aregiven inTable 3.

From (5.20) and Table 3wededu e that

jsin(

1

logd+

2

+argA

d (s

0

))j>R

2 (d

2 (m)):

Inlightof(5.18),(5.7)followsimmediately. Sin e(5.3)isfalse,we on lude

that h( d) 6= m for m 2f5;7;:::;23g and d

2

(m) d  10 850

. Note that

fromTable2inx2,wehave ertainly overedtheranged

2

(m)d d

3 (m).

Table 3. h( d)6=mford

2

(m)d10 850

m d

2 (m) R

2 (d

2

(m)) 

2 (d

2

(m)) 

2

(m) jsin

2 (m)j

2

(m) jsin

2 (m)j

5 10 65

2:2
10 14

1:4
10 7

0:264 0:26 3:121 0:020

7 10 65

3:3
10 14

1:1
10 5

0:264 0:26 3:121 0:020

9 10 66

2:5
10 14

9:9
10 5

0:268 0:26 3:121 0:020

11 10 68

9:9
10 15

3:9
10 4

0:275 0:27 3:121 0:020

13 10 73

6:8
10 16

8:2
10 4

0:293 0:28 3:121 0:020

15 10 76

1:5
10 16

1:7
10 3

0:303 0:29 3:122 0:019

17 10 79

3:1
10 17

2:9
10 3

0:312 0:30 3:124 0:017

19 10 79

3:8
10 17

5:6
10 3

0:310 0:30 3:126 0:015

21 10 79

4:8
10 17

1:1
10 2

0:304 0:29 3:132 0:009

23 10 79

6:8
10 17

2:0
10 2

0:295 0:29 3:141 5:9
10 4

The d

2

(m) listedin Table 3are fartoo largeto allowthe ompletion of

the lass number m problem using the omputationally intensive sieve in

x6 to investigate the range d < d

2

(m). Pushing the pre eding arguments

to their limit, it is possible to produ e a d



2

(m) (< d

2

(m)) su h that if

d



2

(m) < d < d

3

(m), then h( d) 6= m. Approximate values of d



2

(m) are

given in Table 4. Note, however, that for odd m > 9, d



2

(m) is also far

too large to allow one to omplete the lass number problem by using a

omputationallyintensivesieve. Sometypeoffurtherargumentisne essary.

Inx5.2andx5.3, theideaofpartitioningminimaisintrodu ed,whi hallows

ustoredu etherangeofadmissibledis riminantssothata omputationally

intensivesieve anbeused. Itturnsoutthatournewargumentsarepowerful

enoughtoruleouttheranged



2

(m)<d<d

2

(m),obviatingtheneedtopush

theargumentsof thissubse tion to theirtediouslimits.

Table4. Lowerboundsondobtainedwithoutpartitionestimates

m d



2

(m) 4(m 1) 2m

5 4:5
10 12

4:2
10 6

7 4:4
10 14

3:1
10 11

9 2:2
10 18

7:2
10 16

11 7:0
10 23

4:0
10 22

13 8:7
10 29

4:6
10 28

15 2:1
10 36

9:7
10 34

17 8:4
10 42

3:5
10 41

19 5:3
10 49

2:0
10 48

21 5:0
10 56

1:8
10 55

23 6:8
10 63

2:3
10 62

Thelast olumnofTable4unders oresthefa tthatthemethodsinthis

subse tion do not suÆ e when m > 9. This olumn arises from the fa t

that the denominator on the right-hand side of (5.9) must be positive. In

other words, d >4(m 1) 2m

. Note that when m > 9, the last olumn of

Table 4 pre ludes the use of the omputationally intensive sieve in x6, no

matter how sharpwe make the otherestimates inthissubse tion.

5.2. Therange d

1

(m)d d

2 (m)

Lemma10.Let t2R +

. Assumethatm=h( d)isoddandthepartition

P overs M

d

in the senseof Denition 2. Then

(5:21) jA

d

(1=2+it)j1 2 X

a2M



d l

a (d)

1=2

:

Furthermore, if the right-hand sideof (5:21) ispositive, then

(5:22) jargA

d

(1=2+it)j 2t

P

a2M



d f(l

a (d))

1 2 P

a2M

 l

a (d)

1=2

wheref isthe nonin reasingfun tion dened by

f(x)=



2=e for 0xe 2

,

x 1=2

logx for x>e 2

.

Proof. The proofis identi alto theproof of Lemma 8 ex eptthat in-

equality(3.4)isrepla edwiththepartitioninequalitiesl

a

(d)au

a (d).

Lemma 11.Let t2R and m=h( d)>1. Suppose that m is odd and

thepartition P oversM

d

inthesenseofDenition2. If k isanoddprime,

then

jU

d

(1=2+it)j

24m(

p

3k+2)

d 1=4

(

p

3 ) 3=2

k 2

e

 p

3k=2

+ 16

p

k(

p

d+2k)

 3=2

d

e

 p

d=(2k)

+ 16

d 1=4

X

a2M



d F



 p

d

u

a (d)k



:

Proof. We use Lemma9. In the outer sum of (5.15), r =1 orr = k.

By Lemma 7, F is de reasing, so we an boundthe innersummands from

above by usingupper bounds on a. When r =k we usethe general upper

bounda (d=3) 1=2

of (3.4). When r = 1 and a >1 we usethe partition

inequalityau

a

(d). Thus, we have

jU

d

(1=2+it)j



12m

d 1=4

k 1=2

F(

p

3k)+ 8

d 1=4

F



 p

d

k



+ 16

d 1=4

X

a2M



d F



 p

d

u

a (d)k



:

UsingLemma6withN =0andx= p

3kandtheinequalitylog (1+x)<x

(validforx>0), we have

jF(

p

3k)j 2

p

3 1=4

p

k



1+ 2

 p

3k



e

 p

3k=2

:

Similarly,forx= p

d =k,weobtain

jF(

p

d=k)j 2

p

k

p

d 1=4



1+ 2k

 p

d



e

 p

d=(2k)

:

Lemma11 follows easily.

For the remainder of x 5.2, we assume there exists a dis riminant d

satisfyingthefollowingproperties:

(I) h( d)=m,where m2f5;7;9;:::;23g is xed;

(II) d

1

(m) d  d

2

(m), whered

1

(m) is given in Table 5 near the end

of thissubse tionand d (m) isgiven inTable3.

Our goal is to show that (5.7) is true for d with a suitable hoi e of k

and s

0

. Ina mannersimilarto x5.1, we let s

0

=1=2+it

0

and use

k=17923 and t

0

=0:030985799:

Weinberger[31℄hasshownthattheerrorint

0

islessthan10 8

butweonly

usethe rst5 signi ant digits. From (5.4){(5.6) we thenhave



1

=0:01549; 

2

=0:2216; 

3

=57:1;

whereall approximationsarea urateto the numberof pla esshown.

As in x3, let M

d

denote the multiset of minima for d. Turning to

AppendixB,notethatea h entry inTableB m 3

2

(e.g., when m=15,refer

to Table B6) onsists of (m 1)=2 pairs of fun tions (l

a (d);u

a

(d)), ea h

in reasing in d, orresponding to the (m 1)=2 elements in M



d

. In other

words,ea hentryofTableB m 3

2

isapartitionforda ordingtoDenition1

in x3. Sin e d  d

1

(m)  2 m+2

from Table 5, the arguments at the end

of x 3 show that there is some partition for d in Table B m 3

2

whi h overs

M

d

inthe senseof Denition2 in x 3. We hen eforth denote thispartition

byP

1 .

Applyingthe rstpart ofLemma 10 witht=t

0

and P =P

1

,we have

jA

d (s

0

)j1 2 X

(l

a

;u

a )2P

1 l

a (d)

1=2

(5:23)

min

P



1 2 X

(l

a

;u

a )2P

l

a (d)

1=2



;

wheretheminimumishen eforthunderstoodtobeoverallpartitionsP for

d o urringinTableB m 3

2

. Sin el

a

(d)isin reasingind,werepla edwith

d

1

(m)inthe lowerboundof (5.23) givingthe newlowerbound

(5:24) min

P



1 2 X

(la;ua)2P l

a (d

1 (m))

1=2



>0;

withthe positivityfollowingfrom dire t omputation.

From(5.23),(5.24), andLemma11witht=t

0

,k =17923,and P =P

1 ,

we have

jU

d (s

0 )j



3 jA

d (s

0 )j

R

1 (d;P

1 );

wherewedene

R

1

(d;P)=



24m(

p

3k+2)

d 1=4

(

p

3) 3=2

k 2

e

 p

3 k=2

+ 16(

p

d+2k)

 3=2

d e

 p

d=(2k)

+ 16

d 1=4

X

(l ;u )2P F



 p

d

u

a (d)k



n



3



1 2 X

(l ;u )2P l

a (d)

1=2

o

:

Thusfrom (5.23) and (5.24) we have

jU

d (s

0 )j



3 jA

d (s

0 )j

max

P R

1 (d;P);

wherethemaximumishen eforthunderstoodtobeoverallpartitionsP for

d o urringinTableB m 3

2 .

Note that for ea h u

a

o urring in ea h of the partitions for d in

Table B m 3

2 ,

p

d=u

a

(d) is a nonde reasing fun tion of d. It follows from

Lemma 7 that F(

p

d =(u

a

(d)k)) is a nonin reasing fun tion of d. There-

fore, the numeratorof R

1

(d;P) isde reasing ind for any partitionP for d

appearinginTable B m 3

2

. Thus, withtheaidof (5.24), we have

(5:25)

jU

d (s

0 )j



3 jA

d (s

0 )j

max

P R

1 (d

1

(m);P)

sin e d  d

1

(m). Upper bounds for max

P R

1 (d

1

(m);P) an be found in

Table 5. In order to produ e these approximations, we need estimates for

the fun tionsF evaluated at smallpositive arguments. Su h estimates are

easilyobtainedbyapplyingLemma6 withsuitably largeN.

Note that from (5.23) and (5.24) we know the right-hand sideof (5.21)

is positive for all partitions for d appearing in Table B m 3

2

. Applying the

se ond partof Lemma10 witht=t

0

and P =P

1

,we obtain

jargA

d (s

0 )j

1 (d;P

1 );

wherewedene



1

(d;P)= 2t

0 P

(la;ua)2P f(l

a (d))

1 2 P

(l

a

;u

a )2P

l

a (d)

1=2 :

It followsfrom therightmost inequalityin(5.23) and (5.24) that

(5:26) jargA

d (s

0

)jmax

P 

1 (d;P);

wherethemaximumisoverall partitionsP ford o urringinTableB m 3

2 .

Now

1

(d;P) isde reasingindforanyxedpartitionP ford appearing

inTableB m 3

2

. Hen e,

(5:27) jargA

d (s

0

)jmax

P 

1 (d

1

(m);P)

sin edd

1

(m). Table5 ontainsupperboundsformax

P 

1 (d

1

(m);P).

Let 

1

(m) bedened by



1

(m)=

1 logd

1

(m)+

2

max

P 

1 (d

1

(m);P)

and

1

(m)bedened by

1

(m)=

1 logd

2

(m)+

2

+max

1 (10

37

;P):

Lower bounds for

1

(m) and upperboundsfor

1

(m) aregiven inTable 5.

UsingTable5,thefa t thatd

1

(m)d and (5.27), wehave

0<

1

(m)

1

logd+

2

+argA

d (s

0 ):

In theotherdire tion,we laim that



1

logd+

2

+argA

d (s

0 )

1

(m)<:

Note thatfrom Tables3 and 5wehave

d

1

(m)<10 37

<d

2 (m):

If 10 37

<d, our laimfollowsfrom the fa tthat dd

2

(m), (5.26) oupled

withthefa tthat

1

(d;P) isde reasingind foranyxedpartitionP ford

appearinginTableB m 3

2

,andTable5. Whend10 37

,notethatby(5.27)

we have



1

logd+

2

+argA

d (s

0 )

1 log (10

37

)+

2

+max

P 

1 (d

1

(m);P):

Withtheaid ofTable5,dire t al ulation shows that



1 log (10

37

)+

2

+max

P 

1 (d

1

(m);P)<

1 (m):

Thusour laimis alsotrue when d10 37

. Hen e,

jsin(

1

logd+

2

+argA

d (s

0

))jminfjsin

1

(m)j;jsin

1 (m)jg:

From thepre eding inequalityand Table 5,we on ludethat

jsin(

1

logd+

2

+argA

d (s

0

))j>max

P R

1 (d

1

(m);P):

Inlightof(5.25),(5.7)followsimmediately. Sin e(5.3)isfalse,we on lude

thath( d)6=mfor m2f5;7;:::;23g and d

1

(m)dd

2 (m).

Table 5. h( d)6=mford

1

(m)dd

2 (m)

m d

1

(m) max

P R

1 (d

1

(m)) max

P 

1 (d

1

(m);P) max

P 

1 (10

37

;P) 

1 (m)

1 (m)

5 3:6
10 11

0:555 0:041 2:39
10

4

0:592 2:542

7 2:0
10 12

0:547 0:072 1:89
10

3

0:588 2:543

9 7:9
10 12

0:519 0:134 5:70
10

3

0:547 2:583

11 4:6
10 13

0:444 0:245 1:15
10

2

0:463 2:660

13 4:9
10 14

0:291 0:448 1:89
10

2

0:297 2:846

15 1:9
10 16

0:148 0:643 2:77
10

2

0:158 2:961

17 1:2
10 18

0:062 0:794 3:82
10

2

0:072 3:079

19 9:2
10 19

0:024 0:906 5:06
10

2

0:027 3:091

21 7:7
10 21

0:009 0:991 6:58
10

2

0:010 3:107

23 6:6
10 23

0:004 1:065 8:47
10

2

0:005 3:125

Form2f5;7;9;11;13g,thed

1

(m)inTable5issmallenoughtoallowthe

sieveofx6. However,form2f15;17;19;21;2 3g, wehavetoresorttofurther

renementsinx 5.3.

5.3. The range d

0

(m)dd

1 (m)

Lemma12.Let t2R +

. Assumethat m=h( d)isoddandthepartition

P oversM

d

inthe senseof Denition2. Let p bethesmallestnonprin ipal

minimum in M

d

. If l is a nonnegative integer su h that the orresponding

set S

l

=fp;p 2

;:::;p l

gM



d

(where S

0

is understood to bethe empty set),

then

(5:28) jA

d

(1=2+it)j 





 1 2

1 

p (t)

l+1

1 

p (t)







 2

X

a2M



d S

l l

a (d)

1=2

;

where 

p

(t) = (p)p 1=2 it

. Furthermore, if <A

d

(1=2+ i) > 0 for

0 t andthe right-hand sideof (5:28) ispositive, then

(5:29) jargA

d

(1=2+it)j

2t(logpj P

l

j=1 j

j

p (t)j+

P

a f(l

a (d)))





1 2 1 

l+1

p (t)

1 

p (t)





2 P

a l

a (d)

1=2

;

wheref isasdened in Lemma 10 andthe sums areover all a2M



d S

l .

Proof. The proof uses arguments similar to those in the proofs of

Lemmas 8and 10.

For the remainder of x 5.3, we assume there exists a dis riminant d

satisfyingthefollowingproperties:

(I) h( d)=m,where m2f9;11;:::;23g is xed;

(II) d

0

(m) d  d

1

(m), whered

0

(m) is given in Table 6 near the end

of thissubse tionand d

1

(m) isgiven inTable5.

Our goal is to show that (5.7) is true for d with a suitable hoi e of k

and s

0

. Asinx 5.2,let s

0

=1=2+it

0

and use

k =17923 and t

0

=0:030986:

From (5.4){(5.6) we thenhave



1

=0:01549; 

2

=0:2216; 

3

=57:1;

whereall approximationsarea urateto the numberof pla esshown.

For ea h partition P for d in Table B m 3

2

, dene `

P

to be the num-

ber of powers of the smallest minima in M



d

that an be shown to appear

in P using the arguments of x 3, if the number of su h powers ex eeds 2.

Otherwise, set `

P

= 0. Denoting the smallest minimum in M



d

by p, let

S

`

P

=fp;p 2

;:::;p

`P

g if `

P

>0, and the empty set if `

P

=0. Finally, let

P



denotethe setof (l

a (d);u

a

(d)) inP forwhi ha62S

`P .

Sin e d  d

0

(m) > 2 m+2

by Table 6, the arguments at the end of x3

showthereissome partitioninTableB m 3

2

whi h oversM

d

inthesenseof

Denition2inx 3. Wehen eforthdenote thispartitonbyP

0

. Applyingthe

rstpartofLemma12witht=t

0

,P =P

0 ,l=`

P0

,andletting

p

=

p (t

0 ),

we have

jA

d (s

0 )j







 1 2

1 

`

P

0 +1

p

1 

p 





 2

X

(l

a

;u

a )2P



0 l

a (d)

1=2

(5:30)

min

P







 1 2

1 

`

P +1

p

1 

p 





 2

X

(l

a

;u

a )2P

 l

a (d)

1=2



;

wheretheminimumishen eforthunderstoodtobeoverallpartitionsP for

d appearing in Table B m 3

2

. Let 

q

= (q)q 1=2 it

0

and note that l

a (d)

and u

a

(d) arein reasing ind. Notethat we an ndafurtherlowerbound

forthelowerboundin(5.30) byevaluatingtheboundat allpossibleprimes

p. To thisend,let

Q=fq prime:l

p (d

0

(m))q u

p (d

1

(m)); (l

p

;u

p

)2Pg:

Now, anew lowerboundfor(5.30) isgiven by

(5:31) min

P min

q2Q







 1 2

1 

`

P +1

q

1 

q 





 2

X

(l

a

;u

a )2P

 l

a (d

0 (m))

1=2



>0;

withthepositivityfollowingfromadire t omputation. From(5.30),(5.31),

and Lemma11 witht=t

0

,k =17923, and P =P

0

,we have

jU

d (s

0 )j



3 jA

d (s

0 )j

R

0 (d;P

0 );

wherewedene

R

0 (d;P)

= 24m(

p

3 k+2)

d 1=4

(

p

3) 3=2

k 2

e

 p

3k=2

+ 16(

p

d+2k)

 3=2

d e

 p

d=(2k)

+ 16

d 1=4

P

P F

 p

d

u

a (d)k



3 min

q2Q 



1 2 1 

`

P +1

q

1 q 



2 P

P

 l

a (d

0 (m))

1=2

:

Thusfrom (5.30) and (5.31) we have

jU

d (s

0 )j



3 jA

d (s

0 )j

max

P R

0 (d;P);

wherethemaximumishen eforthunderstoodtobeoverallpartitionsP for

d o urringinTableB m 3

.

Note that for ea h u

a

o urring in ea h of the partitions for d in Ta-

bleB m 3

2 ,

p

d=u

a

(d) is anonde reasing fun tionof d. LetP be any of the

partitions for d appearing in Table B m 3

2

. It follows from Lemma 7 that

F(

p

d =(u

a

(d)k)) is a nonin reasing fun tion of d. Therefore, the numer-

ator of R

0

(d;P) is de reasing in d for any partition P for d appearing in

Table B m 3

2

. Hen ewith theaidof (5.31) we have

(5:32)

jU

d (s

0 )j



3 jA

d (s

0 )j

max

P R

0 (d

0

(m);P):

Upper bounds for max

P R

0 (d

0

(m);P) an be found in Table 6. In order

to produ e these approximations, we need estimates for the fun tions F

evaluated at smallpositive arguments. Su h estimates are easily obtained

byapplyingLemma 6 withsuitablylargeN.

Note that from (5.30) and (5.31) we know the right-hand sideof (5.28)

is positive for all partitions ford appearingin Table B m 3

2

. Supposethat

0 t

0 . If`

P

0

=0, thenwe have

jA

d

(1=2+i) 1j2 X

(l

a

;u

a )2P

0 l

a (d)

1=2

max

P 2

X

(la;ua)2P l

a (d

0 (m))

1=2

<1;

bya dire t al ulation. Hen e, <A

d

(1=2+i) >0 when `

P0

=0. On the

otherhand, suppose`

P

0

6=0. It followsfrom TableB m 3

2

and Table5that

the smallest minimum p 2 M



d

satises p  (d

1

(m)=4) 1=6

 7406. Hen e,

0<2t

0

logp<,and itis notdiÆ ultto seethat

<A

d

(1=2+i)

1 2

`

P

0

X

j=1 p

j=2

+

2+2 os (2t

0 logp)

p

2 X

P



0 l

a (d

0 (m))

1=2

min

P min

q2Q



1 2

`

P

X

j=1 q

j=2

+

2+2 os(2t

0 logq)

q

2 X

P

 l

a (d

0 (m))

1=2



:

It then follows by dire t omputation that <A

d

(1=2 + i) > 0 for

0    t

0

. Hen e, we may apply the se ond part of Lemma 12 with

t=t

0

,P =P

0

,and l=`

P0

to obtain

jargA

d (s

0 )j

2t

0 (logpj

P

`P

0

j=1 j

j

p j+

P

(la;ua)2P



0 f(l

a (d)))





1 2 1 

`

P

0 +1

p

1 

p 



2 P

(la;ua)2P

 l

a (d)

1=2 :

Hen e,sin el

a

(d) and u

a

(d) arein reasingin d,we have

jargA

d (s

0 )j

0 (d;P

0 );

wherewedene



0

(d;P)=max

q2Q



2t

0 (logqj

P

`

P

j=1 j

j

q j+

P

(l

a

;u

a )2P

 f(l

a (d)))





1 2 1 

`

P +1

q

1 

q 



2 P

(la;ua)2P

 l

a (d)

1=2



:

It followsfrom (5.30) and (5.31) that

jargA

d (s

0

)jmax

P 

0 (d;P):

Sin e f is nonin reasingand l

p

(d) is in reasing in d, we know 

0 (d;P)

isde reasing ind foranyxed partitionP ford appearinginTable B m 3

2 .

Hen e,

(5:33) jargA

d (s

0

)jmax

P 

0 (d

0

(m);P)

sin edd

0

(m). Table6 ontainsupperboundsformax

P 

0 (d

0

(m);P).

Let 

0

(m) bedened by



0

(m)=

1 logd

0

(m)+

2

max

P 

0 (d

0

(m);P)

and

0

(m)bedened by

0

(m)=

1 logd

1

(m)+

2

+max

P 

0 (d

0

(m);P):

Lower bounds for

0

(m) and upperboundsfor

0

(m) aregiven inTable 6.

UsingTable6,thefa t thatd

0

(m)d,and (5.33), we have

0<

0

(m)

1

logd+

2

+argA

d (s

0 ):

Usingthefa tthat dd

1

(m), (5.33), and Table 6,we have



1

logd+

2

+argA

d (s

0 )

0

(m)<:

Hen e,

jsin(

1

logd+

2

+argA

d (s

0

))jminfjsin

0

(m)j;jsin

0 (m)jg:

From thepre eding inequalityand Table 6,we on ludethat

jsin(

1

logd+

2

+argA

d (s

0

))j>max

P R

0 (d

0

(m);P):

Inlightof(5.32),(5.7)followsimmediately. Sin e(5.3)isfalse,we on lude

thath( d)6=mfor m2f9;11;:::;23g and d (m)dd (m).

Table 6. h( d)6=mford

0

(m)dd

1 (m)

m d

0

(m) max

P R

0 (d

0

(m)) max

P 

0 (d

0

(m);P) 

0 (m)

0 (m)

9 6:4
10 12

0:553 0:089 0:589 0:771

11 2:2
10 13

0:555 0:100 0:597 0:810

13 4:2
10 13

0:556 0:116 0:591 0:862

15 9:4
10 13

0:557 0:128 0:591 0:931

17 1:9
10 14

0:556 0:138 0:592 1:005

19 3:5
10 14

0:555 0:149 0:591 1:083

21 6:5
10 14

0:542 0:173 0:576 1:176

23 10:6
10 14

0:548 0:177 0:580 1:249

A omparison ofTables5and 6 shows thatthegap between d

0

(m) and

d

1

(m)is in reasingrapidlyasm in reases from9 to 23.

6. The low range. In this se tion, we omplete the proof of Theo-

rem 1,thestatement of whi happears inx1. Usingtheresultsof x2and x5

(seeTables5and6inparti ular),itsuÆ estondallnegativefundamental

dis riminants d with h( d)2f5;7;:::;23g su h that d1:1
10 15

.

To this end, we rst onsider the small dis riminants d  7:5
10 6

for

whi h an exhaustive sear h isemployed. Forea h d in thisrange,we om-

puted the lass number by ounting the number of redu ed forms of dis-

riminant d. In other words, we sear hed for integers a, b, and with

0<a<(d=3) 1=2

and =(b 2

d)=(4a) su hthat either a<ba< or

0ba= . Thisstraightforward approa h requiredonly32 minutes on

aCrayC90,renderingfurtheroptimizationunne essary. A ompletelisting

ofthenegativefundamentaldis riminantswithodd lassnumbersm inthe

range 1  m  23 that we found in this sear h is given in Appendix A.

It is worth noting that Buell [6℄ had previously omputed lass numbers

of imaginaryquadrati numberelds ford 4
10 6

,and ourresults agree

perfe tlywith his in thisrange. Furthermore, inre ent unpublishedwork,

Buellhasindependentlyveriedourresultsupto7:5
10 6

,usingourmethod

of separatingminimathatwasintrodu edinx3.

Thelargestvalueofdfoundintheabovesear hwasd=90787. Thus,to

ompletetheproof ofTheorem1weneedto showthat thereis nonegative

fundamental dis riminant dwith odd h( d)23 intherange

(6:1) 7:5
10

6

d1:1
10 15

:

It is infeasible to dire tly he k all d in the range (6.1). Instead we used

partition-typeinformationinthefollowingform.

Lemma 13.If d >8 and h( d)23 isodd, then

(i) ( djp)6=0 for all primes p<d;

1=23

(iii)( djp)= 1forall primesp(d=4) 1=6

withatmostoneex eption;

and

(iv )( djp)= 1forallprimesp(d=4) 1=4

withatmosttwoex eptions.

Proof. Item (i) follows dire tly from (2.1) and item (ii) follows from

Lemmas 1 and 5. If two odd primes p;q  (d=4) 1=6

satisfy ( djp) = 1

and ( djq) = 1, then p;p 2

;p 3

;q;q 2

;q 3

;pq;p 2

q;pq 2

2 M

d

, whi h implies

h( d)25 by Lemmas 1 and 4. Thus, item (iii) is true. Lastly, if three

primesp;q;r (d=4) 1=4

satisfy ( djp) =( djq) =( djr) = 1, then Lem-

mas1and4givep;p 2

;q;q 2

;r;r 2

;pq;pr;qr 2M

d

,whi himpliesh( d)25.

Thisprovesitem(iv).

Now, we an build up a substantially smaller set (than (6.1)) of pos-

sible d by using the Chinese Remainder Theorem on the residue require-

ments impli it in (i){(iv) for a set of small primes. Consider an interval

d

0

d  d

1

. Let p

i

denote the ith prime numberand hoose k su h that

m=8 Q

k

i=2 p

i

>d

1

. Here, m is the Chinese Remainder Theorem modu-

lus when onstru ting integersd based on theve tor ofKrone kersymbols

h( djp

i )i

1ik . Let

S

k

(d)=f~"2f1; 1g k

:~"=h( djp

i )i

1ik

satises(i){(iv)g

and forea h ~"2f1; 1g k

let

D

~

"

=f0d<m:h( djp

i )i

1ik

=~"g:

Tosear hallpossibledind

0

dd

1

weusetheChineseRemainderTheo-

remto onstru tD

~

"

forea h~"2S

k (d

0

). Therequirementthat( djp)= 1

(or ( djp) = 1) implies that d is in one of (p 1)=2 residue lasses mod

p for odd primes p, and in one residue lass mod 8 for p = 2. Thus,

jD

~

"

j= Q

k

i=2 (p

i

1)=2 d

1

=2 k+3

. Ea h d 2 S

~

"2S

k (d0)

D

~

"

is then he ked

usingthene essary onditions(i){(iv)fortheprimesfp

k+1

;:::;p

l

gforlsuit-

ably hosenwherebynodsatisesthe onditions. Iflexists,thenthereisno

fundamental dis riminant d withd

0

dd

1

and h( d)2f1;3;:::;23g.

We use this approa h on (6.1) by dividing it up into 3 subintervals orre-

sponding to k=10;12;13.

First, onsider 2:9
10 13

 d  1:1
10 15

. Take k = 13 so that m =

121700105410884 0 > 1:1
10 15

. The bounds in (ii), (iii) and (iv) applied

to d

0

= 2:9
10 13

are 3, 137, and 1604, respe tively. Therefore, ( dj2) =

( dj3) = 1 and at most one of the 11 primes in fp

3

;:::;p

13

g satises

( djp) = 1. Hen e, jS

13 (d

0

)j = 12 resulting in at most 1:3
10 11

possible

o urren esofd modm. Usingl=56,these were eliminatedin 95minutes

on a Cray C90. As a he k on the a ura y of the omputer program,

we printed out the lastholdout, namely d = 12346195539304 3. Note that

( djp)= 1 forall primespp =263 ex eptforp=19,179 and 263.