A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
M. N. H UXLEY
The rational points close to a curve
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 21, n
o3 (1994), p. 357-375
<http://www.numdam.org/item?id=ASNSP_1994_4_21_3_357_0>
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M.N. HUXLEY
1. - Introduction
Let
F(x)
be a real function twicecontinuously
differentiable for 1 x 2, withwhere C > 4. In two
previous
papers[3, 4]
we considered the number ofinteger points
close to the curvefor M m
2M,
in the sense that theinteger point (m, n)
satisfiesin - f (m)l :5 6( 1/2).
When 6 islarge,
thenexponential
sum methodsgive
an
asymptotic
formula. When 6 issmall,
then we cannotexpect
a nonzero lower bound for the number of solutions of theinequality.
In[4]
we gave two methods offinding
upper bounds:Swinnerton-Dyer’s ingenious
determinantal method[5],
useful when T and M are close in order ofmagnitude,
and adifferencing
iteration like that of van derCorput
forexponential
sums(see
Graham and Kolesnik
[2]).
In this note we consider thecorresponding question
for rational
points (a/q, b/q) satisfying
with denominators in a range
Q
q2Q.
We do notrequire
that thehighest
factor of a, b and q be one, but that each rational
point
that can be writtenas
(a/q, b/q)
withQ q 2Q
is counted onceonly,
however manytriples
ofintegers
a,b,
qgive
therequired
ratios. We can consider thetriple
a, b, q as apoint
of theprojective plane, although
the curve y =f (x)
need not bealgebraic.
Pervenuto alla Redazione il 23 Ottobre 1992.
Bombieri and Pila
[1]
have shown that if there areinfinitely
many q for which alarge
number ofpairs
ofintegers
a and bgive points
on aninfinitely
differentiable curve, then the curve must be an arc of an
algebraic
curve.When 6 is
large,
then theasymptotic
formula for the number of rationalpoints
may be derivedby exponential
sum methods. When 6 issmall,
thenwe cannot
expect
a nonzero lower bound for the number of solutions of theinequality.
When both 6 andQ
aresmall,
then it is better to treat each value of qindividually by
the methods of[4].
In this paper we obtain bounds which are useful when 6 is small and
Q
is
large.
Thedifferencing
method of[4]
does notwork,
because the difference of two rational numbers may have a muchlarger
denominator than either. The determinant method of[4]
and[5]
can beused,
but the rationalpoints
close tothe curve do not form a convex
polygon.
We use the
Vinogradov
order ofmagnitude
notation: theinequality
G « Has M,
Q
or T tends toinfinity,
or 6 tends to zero, means that limsup
is bounded
by
a constant that does not involve6,
M,Q
or T, which we call the order ofmagnitude
constant. Thesymbol O(H)
stands for anyunimportant
term G
satisfying
G « H. The relation G » H is definedsimilarly,
and G ~ Hmeans that G » Hand G « H hold
simultaneously.
We can now state ourresult.
THEOREM 1. Let
F(x)
be areal function
twicecontinuously differentiable for
1 x 2, withLet 6, M,
Q
and T be real with M andQ positive integers,
61 /4
andT > M. We write
Let
f (x)
=T F(x/M),
and let R be the numberof points (x, y)
which can bewritten as
(a/q, b/q)
with a, b, qintegers,
M x 2M,Q
q2Q,
and withThen as either M,
Q
or T tends toinfinity,
we have the boundsand, for
any c >0,
The constants
implied
in(1.6) depend
on ê.The order of
magnitude
of the average number of rationalpoints satisfying
the
inequality
isO(bMQ2), corresponding
to the first term in(1.6).
In thespecial
case
F(x)
=x2,
T =M2,
all thepoints (a/q, a2/q2) with q2 Q
in the range M x 2M lie on the curve. The number of thesepoints
has order ofmagnitude
corresponding
to the second term in(1.6).
Hence the bound(1.6)
is not farfrom best
possible.
However bounds like( 1.1 )
for the third derivative ofF(x)
would rule out thespecial
case y =x2.
It islikely
that such bounds wouldpermit
some reduction of the second term in(1.6), along
the lines of[5]
and[4].
For
completeness
wegive
thecorresponding
result for 8 ~1 /4.
THEOREM 2. Let
F(x)
be a realfunction.
Let6,
M,Q
and T be real with M andQ positive integers,
8 ~1 /4.
Letf (x)
and R bedefined
as in Theorem1. Then
We deduce Theorem 2 from the
following
well-knownelementary
lemma.LEMMA 1. Let
;(Q)
denote theFarey
sequenceof
rational numbersa/q
with q > 1,
(a, q)
= 1. For any interval Jof length
a we havePROOF OF THEOREM 2. If
a/q
is the abscissa of a rationalpoint
countedin R, with
a/q
=e/r
in lowest terms, thenb/q
is a rational numberlying
in aninterval of
length 2S/Q
withrlq.
Hencebr/q
is a rational numberlying
in aninterval of
length
a =2br/Q,
with denominator at most2Q/r. By
Lemma1,
the number of
possibilities
forbr/q
is at mostThe distance from
e/r
to the next rational number with denominator at most2Q
is at least1 /2Qr.
Henceby
Lemma 1again
we havewhere the sum is over rationals
e/r
in their lowest terms in the interval Me/r
2M, the resultrequired.
A bound of this form can also be obtainedby considering
each value of qseparately.
D2. - Determinant methods
As in
[4],
we base theargument
onSwinnerton-Dyer’s
mean value lemma[5].
LEMMA 2. For 0 = to ti ... tr real and
f (x)
a realfunction
r timesdifferentiable
we havefor some ~ depending
on x in the open intervalWe number the solutions
Pi
=(ailqi, bilqi)
inlexicographic order,
thatis,
in order of
b/q increasing,
and forb/q fixed,
in order ofa/q increasing.
Weput
with - 2 9i 2.
We define the determinantsWe write A =
T /M2
as in Theorem 1 for the order ofmagnitude
ofDespite
thenotation,
we do not assume that A is less thanunity.
LEMMA 3.
For i j k
and
for some C
betweenailqi
andak/qk
we havePROOF. We
apply
Lemma 2 withThen on
clearing fractions,
we havewhich can be thrown into the determinant form
(2.2).
To prove(2.1 ),
we notethat
2 for each r, and thatHence
which is
(2.1 ).
LEMMA 4. There are at most
646MQ’
valuesof
r withand at most
values
of
r withfor some ~
between M and 2M.COROLLARY. The number
of
rationalpoints
on the curve which can bewritten as
(a/q, b/q)
withQ
q2Q
is at mostPROOF. The
inequality (2.3)
and Lemma 3imply
and the first bound follows from
Similarly
theinequality (2.5) implies
and since
we have
and the second bound
(2.4)
follows from(2.7).
In the
Corollary
we have 6 = 0.By
Lemma3,
the determinantDr,r+l,r+2
must be a nonzero
integer,
so(2.5)
holds: - infact,
with the lower bound1/2 replaced by
one, so we can omit the factor21~3
in(2.4)
togive
a valid upperbound for R - 2. D
We
join
thepoints Pl , ... , PR
to form an openpolygon.
Unlike the case[4]
ofinteger points
close to a curve, thispolygon
need not be convex unless6 is very small. We call
PrPr+1
a minor side of thepolygon
if neitherPr-1
nor
Pr+2
lies on thestraight
linePr Pr+1.
Amajor
side of thepolygon
consistsof a maximal sequence
PrPr+1 ... Pr+t
of collinear consecutive vertices. The determinant in Lemma 3 is nonzero unlessPi, Pj and Pk
are collinear.When is not zero, then one of the two terms on the
right
of(2.2)
isnumerically
at least one half.Putting i = r, j = r + 1
andj
= r and k = r +1),
we see from Lemma 4 that the number of minor sides isat most
The
difficulty
lies inestimating
the number of rationalpoints
which lie onmajor
sides. The three bounds of Theorem 1 come from(2.8)
inconjunction
with the results of Lemmas
10,
12 and 13.LEMMA 5. A
major
sideof
thepolygon P,
...PR
liesalong
a rational linewhere t, m and n are
integers
withhighest
commonfactor (t,
m,n)
= 1.If Pr
and
Pr+1
lie on the line(2.9),
thenPROOF. We have
for i = r, r + 1,
so t,
m and n are in rationalratios,
and we may take them to beintegers
with no common factor. Since(l, m, n)
= 1, we havefor some
integer Eliminating
n, we haveand
LEMMA 6.
If the
verticesPi, Pj, Pk (in order)
are collinearpoints satisfying (1.3),
thenand
PROOF. Since
PZ, Pj
andP~
arecollinear,
the determinant is zero.In
(2.2)
of Lemma 3 we haveWe take moduli on the left hand
side,
substitute the bound(2.1 )
of Lemma3 on the
right
handside,
cancel the factor(ak/qk -
and estimate thedenominators qr
and the second derivative F" at their minimumvalues,
to obtain the firstinequality (2.10)
of the Lemma. The secondinequality (2.11)
followsimmediately.
DLEMMA 7.
Suppose
that the consecutivepoints Pi,..., PZ+a
lie on amajor
side,
whilstPi+d+1
does not. Then eitheror
so that
PROOF. We
write j
= i + d, k = i + d + 1. The area of thetriangle PiPjPk
is the sum of the areas of the
triangles
for r =i, ... , i
+ d - 1. Eachof these
triangles
has areaHence the determinant has size
In Lemma
3,
we have eitherso that
(2.12) holds,
or forsome ~
giving
which
implies
theremaining inequalities.
LEMMA 8. Let
PiPj
be amajor side,
withand
equation
ix + my + n = 0. Thenand also
PROOF. Since the line contains d + 1 rational
points
with x values at leastm/4Q2
apart, we have theinequalities
If the second
possibility (2.13)
of Lemma 7holds,
thenwhich
gives (2.14).
For the bound(2.15)
wemodify
theproof
of Lemma 7.Let e =
[d/2] > d/3. By (2.11)
of Lemma 6 we haveeither with
r = i,
s = i + e or withr = i + e,
s = i + d. We argue as in Lemma 7 withi, j, k replaced by
r, s, i + d + 1.Again
we haveCorresponding
to(2.13)
we havegiving
the second case of(2.15).
DLEMMA 9. The
major
sides with m = 0 contribute at most368MQ2
rationalpoints.
PROOF. In the case m = 0 we have
(t, n)
= 1, so x =-nil,
y =b/q
withtlq.
Let q = kt. Thenly
lies in an interval oflength 28l/Q,
with denominator k2Q/l. By
Lemma1,
there are between 3 andchoices for
b/q
for each fixedn/t. Also,
lies in an interval oflength
M,and t
2Q,
soby
Lemma 1again,
verticalmajor
sides contribute at mostLEMMA 10. Each individual
major
side contributes at mostrational
points.
For A > 1, themajor
sides with andcontribute in total at most
rational
points,
andfor
B116
themajor
sides withcontribute in total at most
rational
points.
COROLLARY. The number
of
rationalpoints
onmajor
sides is at mostPROOF. We use the notation of Lemma 8. Since d > 2, the number of rational
points
on amajor
side satisfies d + 13d/2,
andby (2.16)
in theproof
of Lemma 8 we have
which
gives
the first assertion(2.17).
For the other bounds weadapt
theargument
of Lemma 4. We number themajor
sidessatisfying (2.18)
from 1 to K, and call theirendpoints with dk = j(k) - i(k).
Weput
Since
i(k + 1) > j (k)
andi(k + 2) > j (k) + 1,
the bound(2.14)
of Lemma 8gives
Thus
The bound
(2.19)
follows when we use(2.22)
for the final values k = K - 1 and K, and theinequality dk
+ 13dk/2.
Similarly,
we number themajor
sidessatisfying (2.20),
and define xk likewise.By (2.15)
of Lemma 8and
by (2.22)
for each k, since
OM2 >
1. Thisgives
the bound(2.21).
TheCorollary
followswhen we
give B
its maximum value1/6
in(2.21),
and add the bound of Lemma 9 for the number of rationalpoints
on verticalmajor
sides. D3. -
Duality
For the
deeper
bounds(1.5)
and(1.6)
of Theorem 1 we need the notion of the functiong(y) complementary
tof (x).
It is convenient to havef (x)
definedfor all x,
with f " (x)
>0, although
weonly
use values of x in the range M to 2M. The functionF(x)
is defined for 1 x 2, where it satisfies( 1.1 ).
Suppose
that( 1.1 )
holds on a range a x,~
with a 1,~3
> 2. For x a,either
F(x)
is notdefined,
or theinequality (1.1)
fails. Define(or redefine) F(x)
for x a as the sum of the first three terms in theTaylor
series about x = a, andsimilarly
for x >~3.
Thisgives
a functionF(x)
defined for all x, twicecontinuously differentiable, satisfying
theinequality ( 1.1 ),
andagreeing
with theoriginal
definition for 1 x 2. We continue toput f (x)
=TF(x/M).
The inverse function
h(y)
off’(x)
is defined for all y. Weput
so that
and
We can
easily
check thatf (x)
is the functioncomplementary
tog(y).
LEMMA 11. On a
straight
line lx + my + n = 0, there are at most twodisjoint
intervals with the property that everypoint
on them hasIf
I is such aninterval, containing
two rationalpoints Pi, Pj satisfying (1.3),
with
then
and
PROOF. Since
f 11 (x)
> 0, thepoints x
withform an
interval,
which may be empty. Thepoints
withform a
subinterval,
whichagain
may be empty. The difference of these sets isone or two intervals on the real line.
Let U,
(xo, yo),
be thepoint
on the curve y =f (x)
where thegradient
is-t/m.
Then xo =h(-.~/m),
so thatIn the one-interval case, the value xo must lie within the
interval,
soIn the two-interval case, the value zo does not lie in either interval.
We use the rational
points Pi
andPj. By
subtractionso, for
some ~
betweenai/qi
and we haveby
the mean value theoremThis
completes
theproof
in the one-interval case.In the two-interval case, the
point
xo does not lie in either interval. A secondapplication
of the mean value theoremgives
Since ~
lies between and whilst xo does not, then either or liesbetween ~
and xo. Thus for P,(a/q, b/q), equal
to eitherPi
orPj,
we have .. _ _
and
Taylor’s
theorem about xogives
for some q. Since
f’(xo) _ -£/m,
andla + mb + nq
= 0, we see thatwhich
completes
theproof
of the Lemma. DWe deduce the second bound
(1.5)
of Theorem 1 from(2.8)
and thefollowing
lemma.LEMMA 12. The number
of
rationalpoints
onmajor
sides isPROOF. For technical reasons we want an upper bound for the
length
Aof a
major
side(measured
in thex-direction).
If A exceeds thebound u
ofLemma
6,
then we argue as in Lemma 8. Let thepoints
on themajor
sidebe
Pa, ... , Pa+d.
Then for e =[d/2]
either or span adistance at most it in the x-direction. Instead of
considering
the wholemajor side,
we take the shorter of the twohalves,
which contains at least(d
+1)/2
rational
points, including
theendpoints.
Hence we can suppose that A /~,provided
that we double the final estimate for the number of rationalpoints.
We use Lemma 9 for
major
sides with m = 0, and Lemma 10 with A = C formajor
sidesWe use Lemma
10,
with a value of B to bechosen,
formajor
sides with m >IlBb.
Let N be the
length
of the interval takenby f’ (x)
for M x2M,
so that N CAM.By (3.2)
of Lemma11, major
sides for which(3.4)
is falsehave
t/m
within a distanceof a value of
f’(x).
Thust/m
lies in an interval oflength
M’ « CAM. Wedivide the
major
sides into blocks for which A A 2A,Q’
m2Q’,
whereA and
Q’
are powers of two; A 1 is allowed.By (3.3)
of Lemma11,
and the bound A 1L, thepoint (2013~/m, n/m)
is close to the dual curve y =g(x),
withBy
Theorem 2 with M,Q replaced by M’, Q’,
and with 6replaced by max(6’, 1/4),
there arechoices of t, m and n in such a block. The number of rational
points
on themajor
side isO(AQ2/Q’),
so that we haveThe block contributes
rational
points,
which sums overQ’
toand then over powers of two, A, in the range
sums to
We balance this with the terms
from Lemmas 9 and 10
by choosing
to
get
and the third term is smaller than the second term, so it may be omitted. D The third bound
(1.6)
of Theorem 1 is deduced from(2.8)
and the last lemma.LEMMA 13. For any - > 0, the
major
sides contributerational
points.
Theimplied
constantsdepend
on E.PROOF. We can assume that
since the result follows from Lemma 12 if
(3.8)
is false. As in Lemma12,
weuse Lemmas 9 and 10 with A = C for
major
sides with m = 0, m >1/JM
orwith
(3.4)
false. These casesgive
the terms of(3.7).
Let N be thelength
ofthe interval taken
by f’(x)
for M x 2M, as in Lemma 12.By (3.2)
ofLemma
11, major
sides for which(3.4)
is false havet/m
within a distanceof a value of
f’(x),
since N >AM/C > 11C,
and(3.9)
holds. Thust/m
liesin an interval of
length
We divide the
major
sides into blocks for which A A2A, Q’
m2Q’,
as in Lemma 12.They correspond
to rational lineslx+my+nz
= 0 whichare close to
being tangents,
and these linescorrespond
to rationalpoints
closeto the dual curve y =
g(x) according
to(3.5).
To set up theiteration,
we notethat -
with A’ =
1/ A.
We suppose that we have an upper bound which is a sum ofterms of the form , ~ , _
2a. The number of rational
points
close to the dual curve has anupper bound which is the sum of the
corresponding
termsMultiplying by
the estimate for the number of rationalpoints
on amajor side,
we haverational
points
onmajor
sides with A and m in these ranges.Using (3.6)
as alower bound for A and an upper bound for
Q’,
we see that 2a > 1, then theexpression (3.10)
is at mostThe number of values of
Q’
isO(log 1/6).
Sincewhen A =
C,
then the number of values of A isHence for 1 > 2a > 1, the sum over
Q’
and Arunning through
powers of twogives
times the maximum summand.
The upper bound for the number of rational
points
close to the dual curvewill contain several terms with different
exponents
a,0, -y
and ~. Which termgives
thelargest
contributiondepends
on the sizesof 6,
A, M andQ.
If all thecontributions of these terms are
O(C8MQ2),
then we choose B = 1. Otherwisewe take the
largest
term of the form(3.11)
and equate it toO(B 1/2C6mQ2),
by choosing
so that the
expressions (2.16)
and(3.11 )
are bothThe bound for the number of rational
points
close to the dual curve willalways
include the contribution(2.8)
of the minor sides. The termO(bMQ2)
has a =
1, (3
=0, ~
=3, ~
= 0, so the term(3.11)
becomesO(C8MQ2 / B),
whichcan be absorbed
by
the first term in(3.8).
The term has a = 0,~3
=1 /3, ~
=1, ~
=1/3,
so(3.10)
becomeswhich sums over powers of two,
A ~
andQ’,
towhich is bounded
by
the second term in the first estimate of(3.8).
Since AM > 1, the term(3.12)
can absorb the term in(2.19)
of Lemma 10.When we use the
Corollary
to Lemma 10 to count the rationalpoints
onmajor
sides of thepolygon
of rationalpoints
close to the dual curve, then theterm
O(C81/2MQ2)
with a =1 /2, {3
=0,1
=5/2 and ~
= 1gives
rational
points
when we sum(3.10)
over A andQ’,
andwhen we choose B > 1. This
gives
a three term upper boundfor the number of rational
points
onmajor sides;
the second estimate follows from the firstby
thegeometric
meaninequality.
This bound
(3.13)
is the firststep
of aniteration,
in which wealways
get the same first and third terms, but the second term
changes
under iterationaccording
toWe put Then
so that
If for some n we
have 120n -1/nt
1, thenby (3.16) 20n - qn -
0. Inparticular,
we
have [20n - qn[ 1/2
for alllarge
n, soby (3.17)
qn -30n
-~ 0. We now deduce thatBn, (from (3.14)) and r-n (from (3.15))
all tend to zero. Theinitial values
00
=1/2, ,Qo
= 0, qo =1/2,
~o =1/2 already have 1200 -1/01
1, butthe step from n = 0 to n = 1 is
exceptional
because ao =1/2
introduces an extralogarithm
power. In(3.13)
we have al -5/8, ~31 - 1/8,
11 =19/8, 1/4,
01 - 3/8,
1/1 1 =5/8, so 1201 - 1/d I
=1/8.
Each further iterationstep
reducesthe exponents,
interchanges
M and N with anexpansion
factor 1 +0(£),
andintroduces a numerical factor. After
O(log 1/£)
steps wehave -/2,
and 6, 0, with a numerical factor
depending
on 6’. The term iswhere the order of
magnitude
constantdepends
on -, and we have usedWe can absorb the first term in
(3.13)
if wereplace
Inby 10/3)
andKn
by min(Kn, -11 /6)
in(3.15).
With this iteration rule we have ~n -19/9.
D
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London Math. Soc. Lecture Notes 126, Cambridge 1991.
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292-296.
[4] M.N. HUXLEY, The integer points close to a curve. Mathematika 36 (1989), 198-215.
[5] H.P.F. SWINNERTON-DYER, The number of lattice points on a convex curve. J. Number
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School of Mathematics
University of Wales College of Cardiff Senghenydd Road
Cardiff, CF2 4YH, UK