1. Main results. In [2], H. Iwaniec and A. S´ark¨ozy considered the following multiplicative hybrid problem. Let N be a natural number large enough, G

LXXI.1 (1995)

On a multiplicative hybrid problem

by

Wen-guang Zhai (Jinan)

1. Main results. In [2], H. Iwaniec and A. S´ark¨ozy considered the following multiplicative hybrid problem. Let N be a natural number large enough, G

and G

be subsets of {N + 1, . . . , 2N }, |G

|  N, |G

|  N . They proved that there exist integers n

, n

, b with n

∈ G

, n

∈ G

, and (1) n

n

= b

+ O(b

log

b).

In this paper, we consider a more general case. Let k ≥ 3 be a fixed integer and N be an integer large enough. G

, . . . , G

are k subsets of {N + 1, . . . , 2N }. Suppose ∆ is a real number satisfying 0 < ∆ ≤ 1/2. Let S

denote the number of solutions to the inequality

k(n

. . . n

)

k ≤ ∆, n

∈ G

, . . . , n

∈ G

. We shall estimate S

. Our main result is

Theorem 1. If T

= |G

| . . . |G

|, then

(2) S

= 2∆T

+ O(T

N

log

N ) for k ≥ 4 and

(3) S

= 2∆T

+ O(N

log

N ).

The constant implied in (2) depends on k.

As an application of Theorem 1 we have immediately

Theorem 2. If |G

|  N, . . . , |G

|  N , then there exist integers n

, . . . , n

, b with n

∈ G

, . . . , n

∈ G

such that

(4) n

. . . n

= b

+ O(b

log

b) (k ≥ 3).

The constant implied in (4) depends on k.

R e m a r k. Theorem 2 of [1] implies our Theorem 2 with a weak log factor in the error term for k ≥ 4. So we use a little different method to get a slightly better result.

[47]

Notations. Throughout this paper, kxk = min{|x − n| | n is an integer}, {x} means the fractional part of x, b(x) = {x} − 1/2 and e(x) = exp(2πix).

|G| stands for the number of elements of G. As usual, d

(n) denotes the number of ways n can be written as a product of t factors.

2. Some lemmas. To complete our proof we need some lemmas.

Lemma 1 (Lemma 1 of [2]). Let A and B be two finite sets of real num- bers, A ⊂ [−X, X], B ⊂ [−Y, Y ]. Then for any complex functions u(x) and v(y), we have

   X

x∈A

X

y∈B

u(x)v(y)e(xy)   

≤ 2π

(1 + XY ) X

x∈A

X

x⁰∈A 2Y |x−x⁰|≤1

|u(x)u(x

)| X

y∈B

X

y⁰∈B 2X|y−y⁰|≤1

|v(y)v(y

)|.

Lemma 2 (Theorem 12.2 of [3]). Suppose t ≥ 2 is an integer and write D

(x) = X

n≤x

d

(n) = xP

(log x) + ∆

(x),

where P

(u) is a polynomial of degree t − 1 in u. Then ∆

(x) = O(x

) with θ

= (t − 1)/(t + 1).

Lemma 3. Suppose t is an integer , 2 ≤ t ≤ k, Q

, . . . , Q

are t subsets of {N + 1, . . . , 2N }, δ is a real number , 0 < δ < 1. Let A(Q

, . . . , Q

; δ) be the number of solutions to the inequality

(5) |(n

. . . n

)

− (n

. . . n

)

| ≤ δ, n

, n

∈ Q

, i = 1, . . . , t.

Then

A(Q

, . . . , Q

; δ)  (N

+ δN

log

N )|Q

| . . . |Q

|.

P r o o f. The inequality (5) implies

(6) |(n

. . . n

) − (n

. . . n

)| ≤ δk(2N )

, n

, n

∈ Q

, i = 1, . . . , t.

For any fixed (n

, . . . , n

), the number of solutions of (6) is (7) S(n

, . . . , n

)

≤ D

(n

. . . n

+ δk(2N )

) − D

(n

. . . n

− δk(2N )

).

For simplicity, we put x

= n

. . . n

, y

= δk(2N )

. Then by Lemma 2

we get

(8) S(n

, . . . , n

)

 (x

+ y

)P

(log(x

+ y

)) − (x

− y

)P

(log(x

− y

)) + O((x

+ y

)

)

= 2y

(P

(log x

) + P

(log x

)) + O((x

+ y

)θ

)

 δN

log

N + N

. So we get

A(Q

, . . . , Q

; δ) ≤ X

¯ ni∈Qi

S(n

, . . . , n

)

 (N

+ δN

log

N )|Q

| . . . |Q

|.

This completes the proof of Lemma 3.

Lemma 4. Suppose k

, k

, N

, N

are natural numbers, δ > 0, and Q

and Q

are subsets of {N

, . . . , 2N

} and {N

, . . . , 2N

} respectively. Let A denote the number of solutions to the inequality

|n

n

− n

n

| ≤ δ, n

, n

∈ Q

, n

, n

∈ Q

. Then

(9) A  N

N

log N

+ δN

N

.

P r o o f. The idea of the proof of Lemma 4 comes from [2]. Given r ≤ 2N

and s ≤ 2N

let V

stand for the number of solutions to (10) |n

n

− n

n

| ≤ δ

in n

, n

∈ Q

, n

, n

∈ Q

such that (n

, n

) = r and (n

, n

) = s.

By (10) we get (11)

    n

n

−

 n

n



  δN

N

. Since the points n

/n

are

1

-spaced, we get (12) V

 (1 + δN

N

N

r

)T

by the Dirichlet box principle, where

T

= |{n

∈ Q

| n

≡ 0 (mod s)}|  N

/s.

Thus we obtain

(13) V

 N

s

+ δN

N

r

s

. Similarly, we have

(14) V

 N

r

+ δN

N

r

s

. So

(15) V

 min(N

/s

, N

/r

) + δN

N

r

s

.

Summing over r and s we complete the proof.

3. Proof of Theorem 1 (k ≥ 4). It is easy to check that [n

+ ∆] − [n

− ∆] =

 1, kn

k ≤ ∆, 0, otherwise.

So we have S

= X

ni∈Gi

([(n

. . . n

)

+ ∆] − [(n

. . . n

)

− ∆]) (16)

= 2∆T

+ X

n_i∈G_i

(b((n

. . . n

)

− ∆) − b((n

. . . n

)

+ ∆)).

It is well known that

(17) b(t) = − X

0<|h|≤H

e(ht) 2πih + O

 min

 1, 1

Hktk



and

(18) min

 1, 1

Hktk



= X

a

e(ht)

with

(19) a

 log H

H , a

 min

 1

|h| , H h



if h 6= 0.

Using (17)–(19), we have

(20) X

ni∈Gi

b((n

. . . n

)

± ∆)

= − X

0<|h|≤H

1 2πih

X

ni∈Gi

e(h(n

. . . n

)

± h∆)

+ O  X

ni∈Gi

min



1, 1

Hk(n

. . . n

)

± ∆k



 T

log H

H +

X

min

 1 h , H

h

   X

n_i∈G_i

e(h(n

. . . n

)

)   . So the problem is now reduced to the estimation of the exponential sum

Σ(h) = X

ni∈Gi

e(h(n

. . . n

)

).

Now suppose k ≥ 4 and t = [k/2], then t ≥ 2. Applying Lemma 1 to the sequences A = {h(n

. . . n

)

| n

∈ G

, i = 1, . . . , t} and B = {(n

. . . n

)

| n

∈ G

, i = t + 1, . . . , k}, we get

(21) Σ(h)  (hN V

V

)

,

where V

= A(G

, . . . , G

; 2

(2N )

h

) and V

= A(G

, . . . , G

; 2

(2N )

h

). V

and V

can then be estimated by Lemma 3. We have (22) V

 (N

+ h

N

log

N )|G

| . . . |G

|

and

(23) V

 (N

+ h

N

log

N )|G

| . . . |G

|.

Combining (20)–(23), we get

(24) X

n_i∈G_i

b((n

. . . n

)

± ∆)

 T

log H

H + H

T

N

+ T

N

log

N.

Choosing H such that the first two terms in (24) are equal, we get

(25) X

ni∈Gi

b((n

. . . n

)

± ∆)

 T

N

log

N + T

N

log

N

 T

N

log

N.

Hence Theorem 1 for the case k ≥ 4 follows from (16) and (25).

4. Proof of Theorem 1 (k = 3). Choosing H = N/ log N , we have

(26) X

ni∈Gi

b((n

n

n

)

± ∆)

= − X

0<|h|≤H

1 2πih

X

ni∈Gi

e(h(n

n

n

)

± h∆)

+ O  X

ni∈Gi

min



1, 1

Hk(n

n

n

)

± ∆k



 T

log H

H + S

+ S

, where

S

=    

X

h≤H

1 h

X

n_i∈G_i

e(h(n

n

n

)

± h∆)     and

S

=    X

h≤H²

a

X

ni∈Gi

e(h(n

n

n

)

± h∆)

.

We only estimate S

and we can estimate S

in the same way.

We have

(27) X

h≤H²

a

X

ni∈Gi

e(h(n

n

n

)

± h∆)

= X

L=2^l≥1

X

L≤h<2L ni∈Gi

a

e(h(n

n

n

)

± h∆)

 X

L=2^l

c(L)    

X

L≤h<2L ni∈Gi

a

e(±h∆)

c(L) e(h(n

n

n

)

)    ,

where c(h) = min(1/h, H/h

).

Now we only need to estimate

Σ(L) = X

L≤h<2L ni∈Gi

a

e(±h∆)

c(L) e(h(n

n

n

)

).

Applying Lemma 1 to the sequences A = {hn

| L ≤ h < 2L, n

∈ G

} and B = {(n

n

)

| n

∈ G

, n

∈ G

}, we get

(28) Σ(L)  (LN V

V

)

,

where

V

= X

x,x⁰∈A 2(2N )^2/3|x−x⁰|≤1

1 and V

= X

y,y⁰∈B 2L(2N )^1/3|y−y⁰|≤1

1.

By Lemma 4 we have

(29) V

 LN log N

and

(30) V

 N

log N + N

/L.

Combining (27)–(30), we obtain

(31) S

 N

log

N.

Similarly,

(32) S

 N

log

N.

Hence Theorem 1 for the case k = 3 follows from (16), (26), (31) and (32).

Acknowledgements. The author would like to thank the referee for his

valuable comments and Prof. H. Iwaniec for his encouragement and help.

References

[1] E. F o u v r y and H. I w a n i e c, Exponential sums with monomials, J. Number Theory 33 (1989), 311–333.

[2] H. I w a n i e c and A. S ´a r k ¨o z y, On a multiplicative hybrid problem, ibid. 26 (1987), 89–95.

[3] E. C. T i t c h m a r s h, The Theory of the Riemann Zeta-function, 2nd ed. (revised by D. R. Heath-Brown), Cambridge Univ. Press, Oxford, 1986.

DEPARTMENT OF MATHEMATICS SHANDONG UNIVERSITY

JINAN 250100, CHINA

Received on 22.10.1993

and in revised form on 25.5.1994 and 13.10.1994 (2555)