• Nie Znaleziono Wyników

# Of particular interest is the general problem whether the difference set A − B = {a − b

N/A
N/A
Protected

Share "Of particular interest is the general problem whether the difference set A − B = {a − b "

Copied!
10
0
0

Pełen tekst

(1)

LXI.2 (1992)

Squares in difference sets

by

Jacek Fabrykowski (Winnipeg, Man.)

1. Introduction. There are many problems in number theory that reduce to searching for squares in specific sequences. For instance, we would like to know whether there are infinitely many squares of type p − 1, where p ranges over primes. Of particular interest is the general problem whether the difference set

A − B = {a − b ; a ∈ A, b ∈ B}

contains squares. Furstenberg [1] and S´ark¨ozy [3] studied the case A = B and gave an affirmative answer under the amazingly general condition that A has a positive upper density. S´ark¨ozy [3] succeeded in proving a quantitative version for thinner sets. The best result (under the weakest assumptions) so far has been established by Pintz, Steiger and Szemer´edi [2] showing that A − A contains infinitely many squares if

A(x) = #{a ∈ A ; a ≤ x} > cx(log x)−(1/12) log log log log x, where c is an absolute positive constant.

Various methods have been employed. Furstenberg used ergodic theory, ark¨ozy applied the circle method together with a combinatorial idea and Pintz, Steiger and Szemer´edi introduced further combinatorial refinements.

In this work we apply a variant of the dispersion method which has a potential to give squares in considerably thinner sets.

Let A = (am) and B = (bn) be finite sequences of complex numbers. In applications A and B will be considered as characteristic functions of finite sets. Our aim is to evaluate the sum

(1) S(A, B) =X X X

m−n=l2

ambnl .

Let νd(k) be the number of solutions to x2≡ k (mod d). Put χc(k) =X

d|c

µ c d



νd(k) and sC(k) = 1 2

X

c≤C

χc(k) .

(2)

Theorem 1. Let C ≥ 2 and am= 0 for m > M . Then

(2) S(A, B) =X X

m>n

ambnsC(m − n) + EC(A, B) , where

(3) EC(A, B)  (C−1/2M + M11/12log M )kAk kBk and

kAk = X

|am|21/2

, kBk = X

|bn|21/2

.

R e m a r k s. Notice that χc(k) is multiplicative in c. If (c, 2k) = 1 we have

(4) χc(k) = k

c

 µ2(c) .

For other c the formula for χc(k) is somewhat complicated but still expressed in terms of the quadratic character.

The error term EC(A, B) can be improved a bit by employing the Fourier technique. Regarding the parameter C it yields the best estimate for the error term EC(A, B) with C = M1/6, however such a choice may not be best for handling the main term. Clearly, for that matter we may want C to be small depending on our knowledge of the distribution of A and B in arithmetic progressions. For example, if we deal with primes ≤ M then we can allow C  (log M )A in view of the Prime Number Theorem of Siegel and Walfisz.

2. Dispersion method. We set s(k) = l if k = l2 and s(k) = 0 otherwise, so

S(A, B) =X X

m>n

ambns(m − n) .

Therefore Theorem 1 reveals that the character sum sC(k) approximates s(k) very well in the sense that the bilinear form

SC(A, B) =X X

m>n

ambnsC(m − n)

is close to S(A, B). To estimate the difference EC(A, B) = S(A, B) − SC(A, B) we shall apply the dispersion method.

Let us introduce S(n) = X

m>n

ams(m − n) , SC(n) = X

m>n

amsC(m − n) , and

SC =X X

m>n

ambnsC(m − n) .

(3)

By Cauchy’s inequality we obtain (5) EC(A, B) =X

n

bn(S(n) − SC(n))  D1/2kBk , where

D =X

n

|S(n) − SC(n)|2.

Squaring out and changing the order of summation we write (6) D = hS, Si − 2 RehS, SCi + hSC, SCi , where

hS, Si =X

n

|S(n)|2, hS, SCi =X

n

S(n)SC(n), hSC, SCi =X

n

|SC(n)|2, with the aim of evaluating each sum separately.

3. Evaluation of hSC, SCi. We have hSC, SCi = X

n

X

m1>n

X

m2>n

am1am2sC(m1− n)sC(m2− n)

= 1 4

X X

c1,c2≤C

X

m1

X

m2

am1am2Jc1c2(m1, m2) , where

Jc1c2(m1, m2) = X

n<min(m1,m2)

χc1(m1− n)χc2(m2− n) .

Since χc(k) is periodic in k of period c we split the summation over n into progressions to modulus c1c2and get

Jc1c2(m1, m2) = X

z (mod c1c2)

χc1(m1− z)χc2(m2− z) min(m1, m2) c1c2

+ O(1)



= min(c1, c2) c1c2

jc1c2(m1− m2) + O(c1c2) , where

(7) jc1c2(k) = X

z (mod c1c2)

χc1(z)χc2(z − k) ,

which resembles the Jacobi sum. Here the error term comes from the trivial estimate

(8) X

z (mod c1c2)

c1(z)χc2(z − k)|  c1c2.

(4)

For exact evaluation of jc1c2(k) we appeal to the formula for the Ramanujan sum

(9) Rc(k) = X

x (mod c)

e kx c



=X

d|c d|k

 c d



giving

(10) χc(k) = 1 c

X

x (mod c)

X y (mod c)

e y(x2− k) c

 .

Hence jc1c2(k)

= 1 c1c2

X X

x1(mod c1) x2(mod c2)

XX y1(mod c1) y2(mod c2)

X

z (mod c1c2)

e y1(x21− z) c1

y2(x22− z + k) c2

 .

Here the innermost sum vanishes unless c1= c2= c say, and y1≡ y2(mod c), in which case we get

jcc(k) = X y (mod c)

X X

x1,x2(mod c)

e y(x21− x22− k) c



= |G(c)|2Rc(k) ,

where G(c) is the Gauss sum

G(c) = X

x (mod c)

e x2 c

 .

For subsequent use we recall the well-known formula

(11) |G(c)|2=

c if 2 - c, 0 if 2 k c, 2c if 4 | c.

Collecting the above evaluations we conclude that (12) hSC, SCi

= 1 4

X

c≤C

c−2|G(c)|2X

m1

X

m2

am1am2min(m1, m2)Rc(m1− m2)

+ O



C4 X

m

|am|2 .

(5)

4. Evaluation of hS, SCi. We have hS, SCi = X

n

X

m1>n

X

m2>n

am1am2s(m1− n)sC(m2− n)

= 1 2

X

c≤C

X

m1

X

m2

am1am2Jc(m1, m2) ,

where

Jc(m1, m2) = X

n<min(m1,m2)

s(m1− n)χc(m2− n)

= X

0<m1−l2<min(m1,m2)

χc(l2+ m2− m1)l

= X

z (mod c)

χc(z2+ m2− m1) X

l≡z (mod c) 0<m1−l2<min(m1,m2)

l

= min(m1, m2)

2c jc(m2− m1) + O(cm1/21 ) , where

(13) jc(k) = X

z (mod c)

χc(z2+ k)

and the error term comes from the trivial estimate

(14) X

z (mod c)

c(z2+ k)|  c .

By (10) and (13) we infer that jc(k) = 1

c X

x (mod c)

X y (mod c)

X

z (mod c)

e y(x2− z2− k) c



= |G(c)|2 c Rc(k) . Collecting the above evaluations we conclude that

(15) hS, SCi

= 1 4

X

c≤C

c−2|G(c)|2X

m1

X

m2

am1am2min(m1, m2)Rc(m2− m1)

+ O



C2 X

m

m1/2|am| X

m

|am|

.

(6)

5. Evaluation of hS, Si. We have hS, Si = 2 ReX X X

n<m2<m1

am1am2s(m1− n)s(m2− n) +X X

l2<m

l2|am|2

= 2 ReX X

m2<m1

am1am2J (m1, m2) + O X

m

m3/2|am|2 , where

J (m1, m2) =X X

n<m2

s(m1− n)s(m2− n) .

To evaluate J (m1, m2) we put n = m1− l12= m2− l22and then u = l1− l2, v = l1+ l2. This is a one-to-one correspondence subject to the following conditions:

U1< u < U2, uv = k , u ≡ v (mod 2) , where U1=

m1

m2, U2=

m1− m2 and k = m1− m2. Hence we obtain

J (m1, m2) = 1 4

X

u

X

v

(v2− u2) = 1 4

X

U1<u<U2

k≡u2(mod 2u)

(k2u−2− u2)

= 1 4

X

U1<u<U2

(k2u−2− u2) 1 2u

X

y (mod 2u)

e y(k − u2) 2u



= X

2U1<cr<2U2

2|cr

(cr)−1 k cr

2

 cr 4



X

x (mod c)

e xk

c +xcr2 4



= H(m1, m2) + I(m1, m2) ,

say, where H(m1, m2) denotes the partial sum restricted by c ≤ C and I(m1, m2) denotes the partial sum restricted by c > C.

First we evaluate H(m1, m2). Given c ≤ C we sum over r getting X

R1<r<R2

2|(2,c)r

r−1 k cr

2

 cr 4

2

e xcr2 4



= X

% (mod 2) 2|(2,c)%

e xc%2 4



X

R1<r<R2

r≡% (mod 2)

r−1 k cr

2

 cr 4

2 ,

where R1 = 2U1c−1 and R2 = 2U2c−1. The innermost sum is approxi- mated by

(7)

1 2

R2

R

R1

 k cr

2

 cr 4

2

 dr

r + O m1

R1



= m2

4 + O

 cm1

m1 m2



and the outer sum is clearly equal to |G(c)|2c−1 (see (11)). This gives H(m1, m2) = 1

4 X

c≤C

c−2|G(c)|2m2Rc(m1− m2) + O

 Cm1

m1 m2

 . Now we proceed to estimate I(m1, m2) by an appeal to the large sieve inequality

(16) X

q≤Q

X a(mod q)

X

m≤M

λme a qm



2

≤ (Q2+ M ) X

m≤M

m|2.

We assume that the sequence A = (am) is supported in the interval 1 ≤ m ≤ M and deduce by partial summation that

X X

m2<m1

am1am2I(m1, m2)

C−1M (log M )2 X

c≤2 M

X x (mod c)

X

m≤M

a0me x cm



X

m≤M

a00me x cm

 with some sequences A0 = (a0m) and A00 = (a00m) with |a0m| ≤ |am| and

|a00m| ≤ |am|. Hence by (16) the above sum is

C−1M2(log M )2 X

m≤M

|am|2. Collecting the above results we conclude that

hS, Si (17)

= 1 4

X

c≤C

c−2|G(c)|2X

m1

X

m2

am1am2min(m1, m2)Rc(m1− m2) + O

(CM3/2+ C−1M2)(log M )2 X

m

|am|2

.

6. Proof of Theorem 1. Conclusion. Inserting (12), (15) and (17) to (6) we find that the main terms cancel out and we are left with the error terms giving

(18) D  (C4M + C2M3/2+ C−1M2)(log M )2 X

m

|am|2 . Finally, by (5) we get (2) with

EC(A, B)  (C2M−1/2+ CM3/4+ C−1/2M )(log M )kAk kBk .

(8)

We shall improve this result slightly by estimating the difference SC(A, B) − SC0(A, B) = 1

2 X

C0<c≤C

X X

m>n

ambnχc(m − n) . Using the results of Section 3 and the large sieve inequality we obtain

|SC(A, B) − SC0(A, B)|2

≤ kBk2X

n

X

C0<c≤C

X

m>n

amχc(m − n)

2

= kBk2 X

C0<c1,c2≤C

X

m1

X

m2

am1am2Jc1c2(m1, m2)

= kBk2 X

C0<c≤C

c−2|G(c)|2X

m1

X

m2

am1am2min(m1, m2)Rc(m1− m2) + O(C4M kAk2kBk2)

(C0−1M2+ C4M )kAk2kBk2. This gives

S(A, B) = SC0(A, B)

+O([(C2M1/2+ CM3/4+ C−1/2M ) log M + C0−1/2M ]kAkkBk) . We take C = M1/6 and get (2) with (3).

7. Further assumptions and results. Our goal is to give a more accessible expression for the main term SC(A, B) in Theorem 1. To this end we impose local conditions on the distribution of squares in the difference set. Suppose the sequences A, B satisfy the asymptotic law

X X

m>n

ambnνd(m − n) = ω(d) X

m>n

ambn+ rd(A, B) ,

where rd(A, B) is considered as an error term and ω(d) is a multiplicative function such that

(19) Z(s) = ζ−1(s)

X

d=1

ω(d)d−s is holomorphic and bounded in Re s ≥ −1/2. We obtain

SC(A, B) = 1 2

X

c≤C

 X

d|c

ω(d)µ c d



X X

m>n

ambn+ FC(A, B) , where

(20) |FC(A, B)| ≤ X

d≤C

Cd−1|rd(A, B)| .

(9)

Furthermore, by contour integration we find that X

c≤C

 X

d|c

ω(d)µ c d



= 1 2πi

1/2+iT

R

1/2−iT

Z(s)Csds

s + O(T−1C1/2log C)

= Z(0) + 1 2πi

−1/2+iT

R

−1/2−iT

Z(s)Csds

s + O(T−1C1/2log C)

= Z(0) + O(C−1/2log T + T−1C1/2log C)

= Z(0) + O(C−1/2log C)

by taking T = C. Hence we conclude the following Theorem 2. Under the above conditions we have S(A, B) = 1

2Z(0)X X

m>n

ambn+ FC(A, B)

+ O(M C−1/2log C + M11/12log M )kAkkBk . 8. An application. To illustrate the asymptotic formula of Theorem 2 we consider the sequences A = B = (Λ(n)), the von Mangoldt function.

By the Generalized Riemann Hypothesis we get X X

n<m≤M

Λ(m)Λ(n)νd(m − n) = ω(d)X X

n<m≤M

Λ(m)Λ(n) + rd(A, B) , where

ω(d) = 1 ϕ2(d)

XX α,β(mod d)

νd(α − β) = 1 and

rd(A, B)  (dM3/2+ d2M )(log M )2. Hence

FC(A, B)  (C2M3/2+ C3M )(log M )2. Corollary. We have

X X

n<m≤M

Λ(m)Λ(n)s(m − n) = 14M2+ O(M23/12log M ) .

R e m a r k. Assuming no Riemann hypothesis one gets the above asymp- totics with the error term O(M2(log M )−A) for any A > 0, and the implied constant depending on A.

(10)

References

[1] H. F u r s t e n b e r g, Ergodic behavior of diagonal measures and a theorem of Szemer´edi on arithmetic progressions, J. Analyse Math. 31 (1977), 204–256.

[2] J. P i n t z, W. L. S t e i g e r and E. S z e m e r ´e d i, On sets of natural numbers whose difference set contains no squares, J. London Math. Soc. (2) 37 (1988), 219–231.

[3] A. S ´a r k ¨o z y, On difference sets of sequences of integers. I , Acta Math. Acad. Sci.

Hungar. 31 (1978), 125–149.

DEPARTMENT OF MATHEMATICS AND ASTRONOMY UNIVERSITY OF MANITOBA

Cytaty

Powiązane dokumenty

[r]

(c) Calculate the number of members of the fitness club who attend neither the aerobics course (A) nor the yoga

Port A is defined to be the origin of a set of coordinate axes and port B is located at the point (70, 30), where distances are measured

It is also known that the test gives a positive result for a rabbit that does not have the disease in 0.1 % of cases.. A rabbit is chosen at random from

Choose one decision of the European Court of Human Rights and critically assess the way in which the Court used the concept of the margin of appreciation.. Discuss the application

Zerwanie umowy jest bardziej efektywnym działaniem niż do- trzymanie jej warunków, jeżeli na skutek takie- go postępowania żadna z umawiających się stron nie znajduje się

For the triangle in Problem 4 compute the length of the altitude through the vertex A.. For the tetrahedron in Problem 7 compute the lenght of the altitude through the

(c) Use the chain rule to find the slope of a tangent line at a point (r, s) of the graph of one of these implicit functions?. You may use the chain rule–but you are not allowed to

[r]

Elastic bunch graph matching (EBGM) 16 , the method used for most of the base landmarking methods in this and our previous paper, can be viewed as an ensemble method as it

4. 52 cards were distributed among four players, 13 each. What is the probability that each player obtained at least one card in clubs?2. 5. What is the probability that the

What is more probable: obtaining a sum of points equal to 7 or rolling the same number twice?.2. A coin was tossed

The author would like to thank the referee for his valuable comments and Prof. Iwaniec for his encouragement

(For the case q = 1, this proof was also given in [11].) In fact, it shows that certain cases of Theorem (3.1) are equivalent to Doob’s results.. We end the section by deriving the

Assume finally that both sides contain a number from the lower and one from the upper half, say x, u from the lower and y, v from the upper... These events are not independent; we

(Since the first author studied a problem closely related to the es- timate of M (ϕ, x) in [4], we do not discuss the case f (n) = ϕ(n) here.) In the second half of the paper we

[4] —, Pad´e approximation for infinite words generated by certain substitutions, and Hankel determinants, in: Number Theory and Its Applications, K. W e n, Some properties of

A partition of a non-negative integer n into h colors is a non-increasing sequence of positive integers in which each positive integer is assigned one of h distinct colors, the order

Let us consider the Banach space L(Ω h ) consisting of all mesh functions y h which are bounded and such that the norm ky h k ∗ defined by (3.1)

He provides a connection between local cohomology modules with respect to an ideal of A generated by a d-sequence and modules of generalized fractions derived from a d- sequence..

[r]

the Letter to the Galatians could have been used by the author(s) of the Fourth Gospel in the context of a polemic related to true sonship and true fatherhood. According to Jesus

Therefore, the paper presents the results of research on the possibility of using as plasticizers of rubber compounds the fractions of heavy pyrolytic oil obtained in the industrial