It is well known that the number of integral solutions (x;y) 2 Z 2 to the Thue inequality jF(x;y)j  m is nite

On ubi Thue inequalities and a result of Mahler

by

Jeffrey Lin Thunder (DeKalb, Ill.)

Introdu tion. Let F(X;Y) 2Z[X;Y℄be a form of degree d3 with

integral oeÆ ientswhi h is irredu ibleover the rationalnumbers Q. It is

well known that the number of integral solutions (x;y) 2 Z 2

to the Thue

inequality jF(x;y)j  m is nite. Moreover, one an estimate the number

N

F

(m) ofsu hsolutions.

Let A(F) denote the area of the region f(x;y) 2 R 2

: jF(x;y)j  1g:

Mahlerin[M℄ approximates N

F

(m)bym 2=d

A(F),getting

(1) jN

F

(m) m

2=d

A(F)j=O(m 1=(d 1)

)

as m ! 1, where the onstant impli it in the O notation depends on F.

More re ently,W. S hmidt showed thatN

F

(m) m 2=d

(d+logm); where

theimplied onstantisabsolute([S℄,Chap. 3,Th. 1C),andlatertheauthor

in[T2℄ proved essentiallythat N

F

(m)dm 2=d

. Seealso [MS1℄,[MS2℄.

Now on the one hand, Mahler's result (1) is better in that m 2=d

A(F)

reallyshouldapproximateN

F

(m). Butontheotherhand,N

F

(m)shouldbe

boundedabovebysomefun tionofm andd independentofthe oeÆ ients

of F, asit is in thelatter two resultsmentioned above. While it is known

that A(F) is bounded above (see [B2℄), the impli it onstant in the error

termof(1)a tuallygrowspolynomiallywiththeheightofF (see[B1℄). The

obvious onje tureisthatN

F

(m)andm 2=d

A(F)shoulddierbyafun tion

dependingonlyon m and d.

The authorin [T1℄ proved just su h a resultin the ase of ubi forms,

i.e., when thedegree d=3:

(2) jN

F

(m) m

2=3

A(F)j=O(1+m 29=44

logm)

as m ! 1, where the onstant impli it in the O notation is absolute.

Besidesthefa tthat theerrortermhere,asa fun tionofm, islarger than

thatin (1), thisresult isla kinginanotherwaywhi hwe nowdes ribe.

1991 Mathemati sSubje tClassi ation: 11D75,11J25.

Resear hpartiallysupportedbyNSAgrantMDA904-95-1-1087.

SupposeG(X;Y)2R[ X;Y℄is aform ofdegree d3with dis riminant

D(G) 6=0: For T 2 GL

2

(R) we get a form G T

(X;Y) = G(T(X;Y)): Two

importantobservationsherearerstthattheprodu tA(G)jD(G)j

1=(d(d 1))

isinvariant under su ha tions, i.e.,

A(G T

)jD(G T

)j

1=(d(d 1))

=A(G)jD(G)j

1=(d(d 1))

(see [B2℄), and se ond that N

F

(m) is invariant under the a tion by T 2

GL

2 (Z).

Now suppose F(X;Y) 2 Z[X;Y℄ is a ubi form with a non-zero dis-

riminant. Su h a form is equivalent under the a tion of GL

2

(R) to ei-

ther XY(X Y) orX(X 2

+Y 2

), dependingon whether it fa tors over R

or not. In parti ular, there are only two possible values for the produ t

A(F)jD(F)j 1=6

(namely 3B(1=3;1=3) and p

3B(1=3;1=3); where B is the

standardbetafun tion). Thus, forsu hformsF,themaintermA(F)m 2=3

in(1)and(2)aboveessentiallyde reasesasafun tionoftheabsolutevalue

of the dis riminantjD(F)j, yetneither of the error terms does. In fa t, as

iswellknown(see [S℄,forexample), the height of F isboundedbelowbya

positivepowerof jD(F)j, sothattheerror terminMahler'sresulta tually

in reasesas afun tionof jD(F)j.

What we areabletoprovehereisthefollowingestimatefor ubi forms

where we not only a hieve the m 1=2

in the error term as in (1) with an

absolute onstant as in (2), but this part of the error term de reases as a

fun tionof jD(F)j.

Theorem. LetF(X;Y)2Z[X;Y℄bea ubi formofdis riminantD(F)

whi hisirredu ibleoverQ andletm1. LetN

F

(m)andA(F)beasabove.

Then

jN

F

(m) m

2=3

A(F)j<9+

2008m 1=2

jD(F)j 1=12

+3156m 1=3

:

Thereisnothingspe ialaboutthe onstants9,2008and3156 appearing

above; they are ertainly notoptimal. We have in luded them rather than

use the  notation to show that the onstants we get are notegregiously

large. Notethatthemainterminourtheoremissmallerthantheerrorterm

ifjD(F)jm 2

, at whi h point the errorterm is m 1=3

. This is not so

badsin eN

F

(m)m 1=3

foranyformwiththe oeÆ ientofX 3

orY 3

1.

In otherwords,themainterminthetheorem is largerthantheerror term

exa tlywhen one would reasonablyexpe tit toapproximateN

F (m).

The proof of the theorem follows the typi al approa h of onsidering

\small", \medium"and\large" solutions. The majorimprovement hereon

past eorts is the fairlylarge lower boundobtained for ertain derivatives

arisingfrom therelated Diophantine approximationproblem. Thisis dealt

with inthe following se tion. Our treatment of \medium"solutions seems

Derivatives, dis riminants and heights. In thisse tion we derive

relationsandboundsfor ertainderivativesandheightsarisingfroma ubi

form F interms ofthedis riminant. We rst xsome notation.

Let F(X;Y) 2 Z[X;Y℄ be a ubi form whi h is irredu ible over Q.

Write

F(X;Y)= 3

Y

i=1 (Æ

i X+

i

Y)=a 3

Y

i=1

(X 

i Y)=b

3

Y

i=1 (Y 

i X);

where

a=Æ

1 Æ

2 Æ

3

; b=

1

2

3

; 

i

=

i

Æ

i

and 

i

= 1

i :

Write

f(X)=a 3

Y

i=1

(X 

i

)=F(X;1) and g(Y)=b 3

Y

i=1 (Y 

i

)=F(1;Y):

TheÆ

i

'sand

i

'sarenotuniquely determinedbyF,of ourse,butthe

i 's,



i

's, a,b,f andg are.

Let

i

=det



Æ

j Æ

l

j

l



; j<l; j;l6=i;

and

= 3

Y

i=1

i :

Then jD(F)j=
2

6=0 (sin e F is irredu ibleover Q). Also, theheight of

F,H(F),satises

H(F)=H(

i

)=H(

i )=

Q

3

i=1 p

jÆ

i j

2

+j

i j

2

ont (F) :

(See[S℄,Chap. 3,Lemma2A.) Notethatthej

i

j's areinvariantunder the

a tionof GL

2

(Z),butH(F) ertainlyis not.

Lemma 1. Let F be a ubi form as above. After possibly applying a

T 2GL

2

(Z),we have

min

1i3 fjf

0

(

i )j;jg

0

(

i )jg

jD(F)j 1=6

3
2 1=3

:

If all the 

i

'sarereal,then

ont (F)H(F)130 1=2

2 1=3

jD(F)j 1=3

;

otherwise

3=2 1=2

Proof. Let L

i

(X;Y)=Æ

i X+

i

Y fori=1;2 and 3. We have

0=det 0

 Æ

1 Æ

2 Æ

3

Æ

1 Æ

2 Æ

3

1

2

3 1

A

=Æ

1

1 Æ

2

2 +Æ

3

3

;

and similarly

0=

1

1

2

2 +

3

3

;

sothat

(3) L

3

(X;Y)

3

=L

2

(X;Y)

2 L

1

(X;Y)

1 :

Also,

(4) jf 0

(

i

)j=ja(

i 

j )(

i 

l )j=

j
j

ja(

j 

l )j

=

j
j

jL

i (1;0)

i j

forall i,where j;l6=i,and similarly

(5) jg

0

(

i )j=

j
j

jL

i (0;1)

i j

:

Werst treat the asewhereallthe

i

'sarereal. LetP betheparallel-

ogram denedby

P =f(x;y)2R 2

:j

i L

i

(x;y)j1 fori=1;2g:

Let

1



2

bethesu essiveminimaofP withrespe ttotheintegerlatti e

Z 2

. We then getabasis fa

1

;a

2 g ofZ

2

thatsatises

max

i=1;2 fj

i L

i (a

1

)jg=

1

; max

i=1;2 fj

i L

i (a

2

)jg=

2 :

Leta=a

1

and supposewithoutlossofgeneralitythatj

1 L

1

(a)j=

1 . Let

b be the greatest integer part of

1 L

1 (a

2 )=(

1 L

1

(a)) and let b=a

2 ba.

Thenfa;bgisabasisforZ 2

,j

1 L

1

(b)j

1

andj

2 L

2

(b)j2

2

. Finally,

afterapplyingasuitableT 2GL

2

(Z)toF (T is given bya andb),wemay

assume thata=(1;0) and b=(0;1).

We on lude from this and (3) that, after applying some T 2 GL

2 (Z)

and possiblyreindexing,we have

j

1 L

1

(1;0)j;j

2 L

2

(1;0)j;j

1 L

1

(0;1)j

1

;

j

3 L

3

(1;0)j2

1

; (6)

j

i L

i

(0;1)ji

2

fori=2and 3.

We needan upperboundfor

2

. ByMinkowski'stheorem,we have



1



2

 4

=j
j:

Also,sin e F isirredu ibleoverQ,

j
jjF(1;0)
j= 3

Y

i=1 jL

i

(1;0)

i j2

3

1

by(6). Thus

(7) 

2

2 1=3

j
j 2=3

:

The rst part of Lemma 1 follows from (4){(7). As for the height, we

have by(6),

j
j ont(F)H(F) p

2 

1 p

5 

2 p

13 

2

= p

130(

1



2 )

2

 p

130 j
j

2

;

sothat theupperboundfortheheight followsfrom (7).

Wenow onsiderthe asewhenthe

i

'sarenotallreal. Wemayassume

without loss of generality that Æ

3

;

3

2 R and that Æ

1

= Æ

2 and

1

=

2 .

Notethat thisimplies

3 L

3

(X;Y)= 2Im(

1 L

1

(X;Y))by(3). We let P

betheparallelogramgiven by

P =f(x;y)2R 2

:jRe(

1 L

1

(x;y))j;jIm(

1 L

1

(x;y))j1g

and let 

1

 

2

be the su essive minima of P. We on lude similarly to

above that, after applyinga suitableT 2GL

2

(Z),wehave

(8)

j

1 L

1

(1;0)j=j

2 L

2

(1;0)j p

2

1

; j

3 L

3

(1;0)j2

1

;

j

1 L

1

(0;1)j=j

2 L

2

(0;1)j p

2

2

; j

3 L

3

(0;1)j2

2 :

Similarlyto above,we get4

3

1

j
j. Also,

4(vol (P)) 1

= 





 det



Re(

1 Æ

1

) Im(

1 Æ

1 )

Re (

1

1

) Im(

1

1 )









=jIm(

1 Æ

1

1

1 )j

= 1

2 





 det



1 Æ

1

1 Æ

1

1

1

1

1









= 1

2 





 det



1 Æ

1

2 Æ

2

1

1

2

2









= j
j

2 :

Thus,



2

 4



1 vol (P)

 j
j

2=3

2 1=3

:

The remainderof thelemma followsfrom thisand (8).

Approximations and a gap argument. In thisse tionweprove two

auxiliary results. They are routine Diophantine approximation fare, but

we have in luded them for ompleteness. We ontinue with the notation

establishedabove.

Lemma 2.Let F bea ubi form asinthe Theorem andsuppose(x;y)2

2







 

i x

y 









4m

jy 3

f 0

(

i )j

for some 1i3: Similarly, if x6=0; then for some i,







 

i y

x 







 4m

jx 3

g 0

(

i )j

:

Proof. Supposey6=0and hooseisothatj

i

(x=y)jissmallest. For

j6=iwe have

2 





 

j x

y 







 





 

j x

y 





 +







 

i x

y 







j

j 

i j:

Thus,

m

jy 3

j

 







F(x;y)

y 3









=jf(x=y)j=jaj 3

Y

j=1 





 

j x

y 







 





 

i x

y 





 jf

0

(

i )j

4

by(4). The proof forx6=0 isthesame with theobvious hanges.

From now on, when we write x=y 2 Q it is assumed that x and y are

relativelyprime integerswith y>0.

Lemma 3. Let  2 C and let A;B;C 2 R with C B >A > 0. The

number ofx=y2Q withj (x=y)jA=(2y 3

) andB yC isno greater

than

1+log

2



logC logA

logB logA



:

Proof. Letx

0

=y

0

;x

1

=y

1

;:::be thedistin trationalnumberssatisfying

thehypothesesofthelemma,arrangedsothaty

0

y

1

:::We laimthat

y

n

A(B=A) 2

n

. This is triviallytrue forn=0. We pro eed byindu tion

on n. We have

1

y

n y

n+1

 





 x

n

y

n x

n+1

y

n+1 







 





 x

n

y

n 







 +







 x

n+1

y

n+1 









 A

2y 3

n +

A

2y 3

n+1

 A

y 3

n

;

so that y

n+1

 A 1

y 2

n

: The laim follows from this and the indu tion hy-

pothesis.

By the laim, ify

n

C,thenC A(B=A) 2

n

. Thisimpliesthat

log

2



logC logA

logB logA



n;

provingthelemma.

Small solutions. Sin e A(F), jD(F)j and N

F

(m) are all invariant

underthea tionof GL (Z),we mayassume withoutlossof generalitythat

ourformsatisesthe on lusionsofLemma1. Spe i ally,weassumefrom

nowon that

(9) jf

0

(

i )j;jg

0

(

i )j

jD(F)j 1=6

2 1=3

3

=

1 jD(F)j

1=6

and that

(10) H(

i

)=H(

i

)=H(F)130 1=2

2 1=3

jD(F)j 1=2

=

2 jD(F)j

1=2

fori=1;2 and3.

Lemma 4. Let N 0

F

(m) denote the number of solutions to jF(x;y)j m

withmax fjxj;jyjgm 1=2

=jD(F)j 1=12

. Then

jN 0

F

(m) m

2=3

A(F)j<9+

375m 1=2

jD(F)j 1=12

:

Proof. Forthetimebeing,letM =m 1=2

=jD(F)j 1=12

. By[D℄ wehave

jN 0

F

(m) B

F

(m)j 9+12M;

whereB

F

(m)denotesthearea of theregion

f(x;y)2R 2

:jF(x;y)jm; maxfjxj;jyjgMg:

LetB 0

F

(m)denote thearea oftheregion

f(x;y)2R 2

:jF(x;y)jm; maxfjxj;jyjg>Mg;

sothat B

F

(m)+B 0

F

(m)=m 2=3

A(F).

By (9),

\

jyj>M

\

jx 

i

yj<4m=(y 2

jf 0

(

i )j)

dxdy= 8m

jf 0

(

i )j

\

jyj>M 1

y 2

dy

=

16m

Mjf 0

(

i )j



16m 1=2

1 jD(F)j

1=12

and similarly

\

jxj>M

\

jy 

i

xj<4m=(x 2

jg 0

(

i )j)

dydx

16m 1=2

1 jD(F)j

1=12

foranyi. Thus,byLemma 2,

B 0

F (m)

6
16m 1=2

1 jD(F)j

1=12

<

363m 1=2

jD(F)j 1=12

:

In on lusion, we have

jN 0

F

(m) m

2=3

A

F

(m)jjN 0

F

(m) B

F

(m)j+jB

F

(m) m

2=3

A

F (m)j

9+ 12m

1=2

1=12 +B

0

F

(m)<9+

375m 1=2

1=12 :

Medium solutions. Forthetime being,xan  amongst 

1

; 

2 and



3

. We will say an (x;y) 2Z 2

with y 6= 0 is losest to  if j (x=y)j is

minimal, i.e., no greater than j

i

(x=y)j for any i. While this does not

ruleout(x;y)being losest to two

i

's, itis losest to atleastone ofthem.

Lemma 5. Let B > 0 and let (x

0

;y

0

) be a solution to jF(x;y)j  m

losest to  with B  y

0

< 2B. The number of solutions (x;y) 2 Z 2

to

jF(x;y)jm losest to withB y<2B and (x;y) not a s alarmultiple

of (x

0

;y

0

) isless than 18m=(B

1

jD(F)j 1=6

). The number of su h solutions

that area s alar multipleof (x

0

;y

0

) is no greater than g=g d(x

0

;y

0 ).

Proof. Writeg(x 0

;y 0

)=(x

0

;y

0

), whereg isthegreatest ommondivi-

sor of x

0 and y

0

. Note that jF(x 0

;y 0

)j mg 3

and (x 0

;y 0

) is losest to .

Leta 0

;b 0

2Zsatisfya 0

y 0

b 0

x 0

=1.

Givenanysolution(x;y)satisfyingthehypotheses,we getauniquepair

a;b2Zwith

x=aa 0

+bx 0

; y=ab 0

+by 0

; and a=y 0

x x

0

y:

Lemma2 gives

jgaj=jx

0 y y

0

xj=yy

0 





 x

0

y

0 x

y 







yy

0



4m

y 3

0 jf

0

()j +

4m

y 3

jf 0

()j



<

9m

Bjf 0

()j :

Further, fora given awe have

B ab 0

y 0



y ab 0

y 0

=b<

2B ab 0

y 0

;

givingnomorethanB=y 0

=gB=y

0

gpossiblevaluesofb. Notethat(x;y)

is a s alar multipleof (x

0

;y

0

) if and only if a =0. Lemma 5 now follows

from (9).

Lemma 6. Thenumber of solutions (x;y) to jF(x;y)jm losest to 

withm 1=2

jD(F)j 1=12

y<16m=(

1

jD(F)j 1=6

) is no greater than

36m 1=2

1 jD(F)j

1=12

+144m 1=3

:

Proof. LetB

0

=m 1=2

=jD(F)j 1=12

andletN beleastsu hthat2 N

B

0



16m=(

1 jD(F)j

1=6

). We will use Lemma 5, ounting those solutions with

2 i

B

0

 y < 2 i+1

B

0

for i = 0;:::;N 1. For some, possibly not all,

i =0;:::;N 1, we have positive integers g

i and x

i

=y

i

2 Q with 2 i

B

0



g

i y

i

<2 i+1

B

0 , jF(x

i

;y

i

)j mg 3

and (x

i

;y

i

) losest to . Setting g

i

=0

forall other i,wegetless than

N 1

X

i=0

18m

2 i

B

0

1 jD(F)j

1=6 +g

i

<

36m

B

0

1 jD(F)j

1=6 +

N 1

X

i=0 g

i

=

36m 1=2

1 jD(F)j

1=12 +

N 1

X

i=0 g

i

possiblesolutions.

Now 2 N

B

0

<32B 2

0

=

1

,so that

N <5 log

2

1 +log

2 B

0

<5 log

2

1

+(21=(2log2))m 1=21

:

Thus,

X

g

i

4 1=3

1 m

2=7 g

i

<(5 log

2

1

+(21=(2log2))m 1=21

)4 1=3

1 m

2=7

(5 log

2

1

+(21=(2log2)))4 1=3

1 m

1=3

<138m 1=3

:

It remains to estimate the sum over those g

i

greater than 4 1=3

1 m

2=7

.

Ifg

i

>4 1=3

1 m

2=7

;then we have







 

x

i

y

i 









4m

(g

i y

i )

3

1 jD(F)j

1=6

<

m 1=7

16y 3

i

jD(F)j 1=6

:

Also,sin eg 3

i

mjF(x

i

;y

i

)j1,wehaveg

i

m 1=3

and y

i

B

0 m

1=3

=

m 1=6

=jD(F)j 1=12

:WenowuseLemma3,withA=m 1=7

=(8jD(F)j 1=6

),B =

m 1=6

=jD(F)j 1=12

and C =16m=(

1 jD(F)j

1=6

). A ording to Lemma 3, we

have nogreater than

1+log

2



log

2

C log

2 A

log

2

B log

2 A



1+log

2



7+(6=7)log

2

m log

2

1

3+(1=42)log

2 m



<1+log

2

((7=3)+36 (1=3)log

2

1 )<7

possiblex

i

=y

i

with g

i

>4 1=3

1 m

2=7

. Thus,

X

g

i

>4 1=3

1 m

2=7 g

i

6m 1=3

:

Alltold,we have less than

36m 1=2

1 jD(F)j

1=12 +

N 1

X

i=0 g

i

<

36m 1=2

1 jD(F)j

1=12

+144m 1=3

possiblesolutions.

Lemma 7. Let N 00

F

(m) denote the number of solutions (x;y) 2 Z 2

to

jF(x;y)jm with

m 1=2

jD(F)j 1=12

<maxfjxj;jyjg<

16m

jD(F)j 1=6

:

Then

N 00

F

(m)<12



36m 1=2

1 jD(F)j

1=12

+144m 1=3



<

1633m 1=2

jD(F)j 1=12

+1728m 1=3

:

Proof. Multiplying bytwo the estimate in Lemma 6 above ounts all

those solutionswith

m 1=2

jD(F)j 1=12

jyj<16m=(

1 jD(F)j

1=6

)

thatare losestto. Multiplyingby3takes areofthedierentpossibilities

for. Finally,theargumentsaboveareentirelysymmetri withrespe ttox

and y,i.e., thesame estimates holdfor ountingsolutions (x;y) withx6=0

and y=x losest to  =

i

forsome i.

Large solutions

Lemma8.Fix anibetween 1and 3andsuppose(x;y)2Z 2

isasolution

tojF(x;y)jm losestto

i

withy16m=(

1 jD(F)j

1=6

). Thenthereisan

x 0

=y 0

2Q with (x;y) =g(x 0

;y 0

), where g=g d(x;y) and (x 0

;y 0

) is losest

to 

i

. Further,







 

i x

0

y 0











4m

(gy 0

) 3

1 jD(F)j

1=6

; y

0



16m

1

gjD(F)j 1=6

and gm 1=3

:

Proof. The rst statement is lear given our onventions mentioned

above about writing elements of Q. Certainly x 0

=y 0

is losest to 

i , so

thatthe se ond statement followsfrom Lemma 1and (9). Finally,we have

g 3

g 3

jF(x 0

;y 0

)j=jF(x;y)jm.

For thenext two lemmas, x an  as inthe last se tion and a positive

integer gm 1=3

.

Lemma 9.Let

C=maxf(8m=(g 3

1 jD(F)j

1=6

)) 4

;(8 3

H()) 2
54

3

g:

Thenumber of x=y2Q with16m=(

1

gjD(F)j 1=6

)yC satisfying







 

x

y 









4m

y 3

g 3

1 jD(F)j

1=6

is lessthan

25:6+log

2



4 log

2

1 +

log

2 m

1+2log

2 g



:

Proof. Let

A=

8m

g 3

jD(F)j 1=6

and B =max f2g 2

A;1g:

Note thatby(10),

(11) C maxfA

4

;( 2

2 8

6

jD(F)j) 54

3

g:

We are then ounting the number of x=y 2 Q with B  y  C satisfying

j (x=y)jA=(2y 3

). Weestimatethenumberofsu hx=yusingLemma3.

First supposeB >1,so that log

2

B log

2

A =1+2log

2

g and 16m>

1

gjD(F)j 1=6

. IfC =A 4

, then

log

2

C log

2 A

log

2

B log

2 A

= 3log

2 A

1+2log

2 g

9 3log

2

1 +

3log

2 m

1+2log

2 g

;

sin eA8m=

1

. If C>A 4

,thenby(11),

log

2

C54 3

(2log

2

2

+18+log

2

jD(F)j)

54 3

(2log

2

2

+18+6(4 log

2

1 +log

2 m)):

Thus,

log

2

C log

2 A

log

2

B log

2 A

1+ log

2 C

1+2log

2 g

54 3



2log

2

2

6log

2

1

+43+ 6log

2 m

1+2log

2 g



:

Now supposeB =1. Then A1=(2g 2

) and

log

2

C 54 3

(2log

2

2

+18+log

2

jD(F)j)

by(11). We also have jD(F)j16;sin ejD(F)j2Zand

116m=(g

1

jD(F)j 1=6

)16=jD(F)j:

We get

log

2 C

log

2

jD(F)j

<54 3

(2log

2

2 +19)

and

log

2

jD(F)j

6log

2 A

= log

2

(8m=(

1 g

3

)) log

2 A

log

2 A

1+

3+log

2

m log

2

1

log

2 A

4 log

2

1 +

log

2 m

1+2log

2 g

:

Thus,

log

2

C log

2 A

log

2

B log

2 A

=1+ log

2 C

log

2 A

<6
54 3

(2log

2

2 +19)



4 log

2

1 +

log

2 m

1+2log g



:

In all ases we have

log

2

C log

2 A

log

2

B log

2 A

<6
54 3

(2log

2

2 +19)



4 log

2

1 +

log

2 m

1+2log

2 g



;

sothat

1+log

2



log

2

C log

2 A

log

2

B log

2 A



<25:6+log

2



4 log

2

1 +

log

2 m

1+2log

2 g



:

Lemma9 follows from thisand Lemma 3.

Lemma10.LetCbeasinLemma 9:Thenumber ofx=y2Q withy>C

and yjxj satisfying







 

x

y 









4m

y 3

g 3

1 jD(F)j

1=6

is no greater than 25.

Proof. For x=y 2 Q, set H(x=y) = p

x 2

+y 2

(re all our onventions

about x=y 2 Q) . In parti ular, for the x=y 2 Q onsidered in Lemma 10,

yH(x;y)<2y.

By Theorem 6A of [S℄, Chapter 2 (usingm =2 and  =1=9), there is

some B >0 su h thatall x=y2Q satisfying







 

x

y 







<

1

H(x=y) 10

p

6 =9

haveeitherH(x=y)<(8 3

H()) 2
54

3

orB H(x=y)<B 4
54

3

. Notethatall

x=y2Q satisfyingthehypotheses ofLemma 10 have







 

x

y 









4m

(yg) 3

1 jD(F)j

1=6

<

4m

y 11=4

C 1=4

g 3

1 jD(F)j

1=6

 1

2y 11=4

<

1

H(x=y) 10

p

6=9

and H(x=y)>(8 3

H()) 2
54

3

.

A standard gap argument (Lemma 8B of [S℄, Chapter 2, for example)

shows thatthe numberof x=y2Q with







 

x

y 







<

1

2y 11=4

and B y<B 4
54

3

isno greaterthan

1+

log(log (B 4
54

3

)=logB)

log(7=4)

<25

Lemma 11. Let N 000

F

(m) denote the number of solutions (x;y) 2 Z 2

to

jF(x;y)jm with

max fjxj;jyjg

16m

1 jD(F)j

1=6 :

Then N 000

F

(m)<1428m 1=3

:

Proof. Fix an i between 1 and 3. By Lemmas 8{10, the number of

solutions losest to 

i with

max fjxj;jyjg=y

16m

1 jD(F)j

1=6

isless than

X

gm 1=3

50:6+log

2



4 log

2

1 +

log

2 m

1+2log

2 g



:

Now

X

gm 1=6

50:6+log

2



4 log

2

1 +

log

2 m

1+2log

2 g



m 1=6

(50:6+log

2

(4 log

2

1 +log

2 m))

<m 1=6

(50:6+log

2

(8 2log

2

1 )+log

2 m)

<m 1=6

(54:4+8:7m 1=6

)<64m 1=3

and

X

m 1=6

<gm 1=3

50:6+log

2



4 log

2

1 +

log

2 m

1+2log

2 g



<m 1=3

(50:6+log

2

(4 log

2

1

+3))<55m 1=3

:

Thus,thenumberofsolutions losest to 

i with

max fjxj;jyjg=y>

16m

1 jD(F)j

1=6

islessthan 119m 1=3

. Moreover, theexa t same estimates holdbythesym-

metry ofour argument forthenumberofsolutions losest to 

i with

maxfjxj;jyjg=x>

16m

1 jD(F)j

1=6 :

Arguingexa tly asintheproof ofLemma 7,we get

N 000

F

(m)<12
119m 1=3

=1428m 1=3

:

Proof of Theorem. ByLemmas 7 and 11 we have

N

F

(m) N 0

F

(m)=N 00

F

(m)+N 000

F (m)<

1633m 1=2

1=12

+3156m 1=3

:

Also,

jN

F

(m) m

2=3

A(F)jjN

F

(m) N

0

F

(m)j+jN 0

F

(m) m

2=3

A(F)j

jN

F

(m) N

0

F

(m)j+9+

375m 1=2

jD(F)j 1=12

byLemma4. The Theoremfollows.

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[B1℄ M. Bean,Bounds for the number of solutionsof the Thueequation, M.thesis,

Univ.ofWaterloo,1988.

[B2℄ |,An isoparametri inequality for the area of plane regions dened by binary

forms,CompositioMath.92(1994),115{131.

[D℄ H.Davenport, On a prin iple of Lips hitz, J. LondonMath. So . 26(1951),

179{183.

[M℄ K.Mahler,ZurApproximationalgebrais her ZahlenIII,A taMath.62(1934),

91{166.

[MS1℄ J.Mueller and W. M. S hmidt, Trinomial Thue equations and inequalities,

J.ReineAngew.Math.379(1987),76{99.

[MS2℄ |,|,ThegeneralizedThueinequality,CompositioMath.96(1995),331{344.

[S℄ W.S hmidt,DiophantineApproximation,Le tureNotesinMath.1467,Sprin-

ger,NewYork,1991.

[T1℄ J.L.Thunder,Thenumberofsolutionsto ubi Thueinequalities,A taArith.

66(1994),237{243.

[T2℄ |, On Thue inequalities and a onje ture of S hmidt, J. NumberTheory 52

(1995),319{328.

Mathemati alS ien esDepartment

NorthernIllinoisUniversity

DeKalb,Illinois60115

U.S.A.

E-mail:jthundermath.niu.edu

Re eivedon7.4.1997 (3162)