A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
N IGEL W ATT
An elementary treatment of a general diophantine problem
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 15, n
o4 (1988), p. 603-614
<http://www.numdam.org/item?id=ASNSP_1988_4_15_4_603_0>
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http://www.numdam.org/
NIGEL WATT
1. - Introduction
This paper is concerned with the order of
magnitude
of~8 ( H, ~ ) ,
thenumber of
integer
solutions( h 1, ~ ~ ~ , h4 , k1, ... , ,~4)
ofand
with
where H is a
positive integer,
A > 0 and g is a fixedcomplex-valued
functiondefined on the interval
[ 1, 2).
Thefollowing
result is obtained.THEOREM.
If
g isanalytic
on some open subset Vof
Ccontaining
theinterval
[1,2]
andthen
Interest in
18 (H, A)
goes back to1985,
when E. Bombieri and H. Iwaniecshowed,
in[2],
that ifg(l) (x) = xp,
for 1 x 2, whereQ
is some realPervenuto alla Redazione il 10
Giugno
1988.constant other than 0 or 1, then
Later I obtained the bound
(5)
forQ = 1 (see [5,
Theorem1]).
The bounds
for = 1
areimportant
inapplications
toexponential
sums.In
[1]
Bombieri and Iwaniec used the bound(6),
for{3
-A,
to show that~
(] + it)
«etS+9/56,
for t > E > 0. The bound(5), for Q - 1,
has beenapplied
in[3]
and[4].
To prove
(6)
Bombieri and Iwaniec treated18 (H, A)
as a mean-value of anexponential
sum and used aningenious argument involving
Poisson summation.In this paper the more
elementary techniques
of[5]
areadapted
togeneralize [5,
Theorem1].
Theresulting
Theorem(above)
does notgeneralize
the bound(6)
of Bombieri andIwaniec,
as the condition(4)
is not satisfied forQ
= 2.My
thanks are due to the Science andEngineering
ResearchCouncil, University College
Cardiff and mysupervisor,
Dr M. N.Huxley.
2. -
Preliminary
results and definitionsThe
positive
number(2~20131)~,
which we shall henceforth refer to as a, occursnaturally
in the next Lemma. It satisfies theequation
Now
so
Also
so the sequence
fall)
isgenerated by
the recurrenceformula,
Starting
from the lower bounds0, 1,
9 1 and21
4 fora - 2 , a -1, a °
and a(respec-
tively),
we use the recurrence formula toproduce
the lower boundsfor
a2, ~ ~ ~ , a9 (respectively).
LEMMA 1. For r =
1,...,4,
letThen, for
and, for
PROOF. It is sufficient to prove these results for 6 >
0,
since the case6 = 0 then follows
by letting 6
tend to zero from above.For r =
1, 2, 3, ..., let
Let
Ego, E4
begiven by
thepolynomial expansion,
Define
Fo, ... , ~’4 similarly,
but in terms of y1, ~ ~ ~ , y4 . Thenfor r =
1,...,4. Therefore,
for R =1, ~ ~ ~ , 4,
Starting
from the fact thatEo
=Fo
=1,
we deduce thatand
Let i be an
integer
with 1 i 4. Thenand
Therefore
which is the result
(10).
For r=5,6,7,~~~,
,Therefore,
for R =5,6,7, ... ,
,Now,
for r =1,2,3, ...,
, letUr
begiven by
Then,
forJS = 5,6, 7,
,Starting
from the upper bound of2
for andU4,
weapply
thisinequality
toproduce
the upper boundsfor
U5 , ~ ~ ~ , (respectively). Using (9)
we caneasily
check thatNow suppose that R is an
integer greater
than 9 and such thatThen
by (8).
The result(11),
for r >5,
followsby
induction. Tocomplete
theproof
note that 25
( Us - 21
25 x 162a~, by (9).
Let
k4)
be aninteger
solution of(1).
Then so is( h 1 + t, ~ ~ ~ , k4 + t)
Ifor any
integer
t. As in[5]
we call this set ofinteger
solutions of(1)
afamily f (say).
Theinteger
solution( h 1, ~ ~ ~ k4)
of(1)
is called trivial if andonly
ifh4 )
is apermutation
of( l~1, ~ ~ ~ k4).
Thefamily f
is called trivial if andonly
iffor i= 1,...,4.
LEMMA 2. An
integer
solutionof (1)
is trivialif
andonly if
it is a memberof a
trivialfamily.
The numberof
trivialinteger
solutionsof (1)
and(3)
isPROOF. This Lemma is almost a restatement of
[5,
Lemma4],
since there is a one-to-onecorrespondence
between the trivialinteger
solutions of(1)
and(3)
and the trivialinteger
solutions of(1)
with 0H,
for i =1, ...
4.The
correspondence
isgiven by
LEMMA 3. The
integer
solutionsof (1)
and(3) fall
intoO(H4) families.
PROOF. This Lemma follows
directly
from[5,
Lemma6].
LEMMA 4. Let 6 > 0 and K > 0. Let F be the number
of
non-trivialfamilies
which contain a member(a 1, ... ,
a4,bl, ~ ~ ~ , b4)
withand
Then
PROOF. If K
2,
thenand it follows from
(1)
that there are no such non-trivial families. The rest of theproof
is almost the same as that of[5,
Lemma9], except
that theappeals
to
[5,
Lemma8]
are made with r =1,
a~and yj
=for j
=1, ... 4.
For J
c [H, 2H),
letS(J)
be the set ofinteger
solutions of(1)
and(2)
with and for i =
1, ...4.
LEMMA 5.
If H, 2H)
=J1
U ’ " UJQ,
thenPROOF. The condition
(2) implies
thatand
Conversely
we note that if -A x, yA, then
Therefore theLemma follows
by [4,
Lemma 2.2 and Lemma2.3].
3. - Proof of the Theorem For cëC and r >
0,
letFor
1x2,
letThen R is a continuous
positive-valued
function on~1, 2;1,
so there exists apositive constant p
such thatLet A be the union of the sets with 1 x 2. Then A is a
compact
subset of C. Let B be the maximun value attainedby Ig(z)1 I
forBy (4),
B > 0. If 1 x
2,
thenby Cauchy’s Integral Formulae,
For
1 x 2,
letBy (4),
E is apositive-valued
continuous functionon 1, 2~,
so there exists apositive
constant e such thatBy (13),
Let
Q
be the leastinteger
withWe
divide [H,2H)
intodisjoint
intervalsJi,
ofequal length.
Let Jbe one of these intervals. Let
( h 1, ~ ~ ~ ,
nS ( J) ,
wheref
is afamily.
Let(ai, , 64)
be the member off
withb 4
= 0.Then,
for i =1,...,4,
so that
For
r=0,1,2,~~~, let
Let
By (1), (17)
and Lemma 1(11),
and
To
estimate f n S (J) ) I
we consider the functionwhich
(by (12))
iscertainly analytic for I z
p. The members off
nS(J)
arein one-to-one
correspondence
with thoseinteger
solutions t ofwhich have
Now, by (7), (15)
and(16),
so that
Therefore,
for n =0, 1, 2, ~ ~ ~,
,where cr =
~4 , for
r =0 1, 2, ~ ~ .
. The various( ~ )
have similarTaylor
series.Combining
these results we find thatHence, by (13), (19), (20)
and(23),
Now,
if n and s arenon-negative integers
and 0 x1,
thenby Taylor’s
Theorem there with
Therefore,
if we let ,then,
forn=0,1,2,~~~,
,by (16).
Now, by (23),
so
Therefore,
sinceit follows from
(25) that,
for r =3, 4,
Hence, by (13)
and(14),
where
By (23), pQ
>1,
so it now follows from(18)
thatand
We now return to
(25)
andapply
the results(13), (15), (26), (27)
and(28)
to showthat,
for n =0,1, 2, ~ ~ ~ ,
INow suppose that
(21)
and(22)
hold.Then, by (23),
By (7), (16), (23), (24), (26), (29)
and(30).
Therefore, by (31 ),
and
Hence, by (2), (21), (23)
and(30),
Therefore, by (27).
so
We can now bound First note
that, by (17)
and(18),
Therefore,
it is sufficient to consider thefollowing
three cases.Case 1 :
By integer
Lemma 1 solution(10), f
of(1)
is and a trivialS(J)
containsfamily. By
at most Lemma24H4 2, (h1,...,
such solutions.lc4)
is a trivialCase 2:
By (32),
Hence, by
Lemma3,
at mostO ( H4 )
members ofS ( J)
fall into Case 2.Case 3:
By
Lemma2, f
is not a trivialfamily.
Since 01,b4
areintegers
and iit follows from
(17)
and(18)
thatWe now consider the
O(log H) non-empty
subcases of theform,
where
A2
is aninteger
power of 2.By (32),
By (17), (18)
and Lemma4,
at mostO(02H41og2H)
families fall into any one such subcase.Hence,
at most members ofS(J)
fall into each subcase and the total numberfalling
into Case 3 is thereforeCollecting
the results from the three cases we find thatThe Theorem now follows
by
Lemma 5.REFERENCES
[1]
E. BOMBIERI - H. IWANIEC, On theorder of 03B6 (1/2 +it) ,
Ann. Scuola Norm.Sup.
PisaC1. Sci. (4) 13 (1986) 449-472.
[2]
E. BOMBIERI - H. IWANIEC, Some mean-value theoremsfor exponential
sums, Ann.Scuola Norm.
Sup.
Pisa Cl. Sci. (4) 13 (1986) 473-486.[3]
M.N. HUXLEY - N. WATT,Exponential
sums and the Riemannzeta-function,
Proc.London Math. Soc. (3) 57 (1988) 1-24.
[4]
N. WATT,Exponential
sums and the Riemannzeta-function
II, J. London Math. Soc.,to appear.
[5]
N. WATT, Aproblem
on semicubical powers, Acta Arith., to appear.Department
of Mathematics, Hill Center, BuschCampus, Rutgers University,
New Brunswick NJ 08903