On a diophantine inequality involving prime numbers

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1 ACTA ARITHMETICA LXI.3 (992 On a diophantine inequality involving prime numbers by D. I. Tolev (Plovdiv In 952 Piatetski-Shapiro [4] considered the following analogue of the Goldbach Waring problem. Assume that c > is not integer and let ε be a positive number. If r is a sufficiently large integer (depending only on c then the inequality ( p c + p c p c r N < ε has a solution in prime numbers p,..., p r for sufficiently large N. More precisely, if the least r such that ( has a solution in prime numbers for every ε > 0 and N > N 0 (c, ε is denoted by H(c then it is proved in [4] that lim sup c H(c c log c 4. Piatetski-Shapiro also proved that if < c < 3/2 then H(c 5. One can conjecture that if c is near to unity then H(c 3. On the other hand, instead of fixed ε, we may consider ε depending on N and tending to zero as N tends to infinity. This conjecture is proved in [7] for < c < 27/26 and ε = N (/c(27/26 c log 3 N. In this paper we sharpen the last result and prove the following theorem: Theorem. Let < c < 5/4. There exists a number N 0 (c > 0 such that for each real number N > N 0 (c the inequality p c + p c 2 + p c 3 N < N (/c(5/4 c log 9 N has a solution in prime numbers p, p 2, p 3. Notation c fixed real number such that < c < 5/4, N sufficiently large number,

2 290 D. I. Tolev (2 (3 (4 (5 X = (N/2 /c, τ = X 3/28 c/2, T = X 4/56+c/4, ε = X (5/4 c log 8 X, (6 = 0 ε, (7 r = [log X], [α] denotes the integer part of the real number α; (8 K = X 5/4 c (log X 6, (9 (0 ( (2 m, n, k, l, d, r (with or without subscripts integers, p, p, p 2,... prime numbers, τ(n the number of positive divisors of n, Λ(n von Mangoldt s function, Ψ(x = n x Λ(n Chebyshev s function, ϱ = β + iγ non-trivial zero of the Riemann zeta function ζ(s, sum over the non-trivial zeroes of ζ(s such that a < γ < b, a<γ<b e(x = e 2πix, A B means A B A, S(x = I(x = I ϱ (x = X/2<p X X/2 X/2 J(x = γ T e(t c x dt, log p e(p c x, e(t c xt ϱ dt, I ϱ (x. The constants in O-terms and -symbols are absolute or depend only on c. To prove the theorem we need some lemmas. Lemma. Let a, δ be real numbers, 0 < δ < a/4, and let k be an integer. There exists a function ϕ(y which is k times continuously differentiable and such that ϕ(y = for y a δ, 0 < ϕ(y < ϕ(y = 0 for a δ < y < a + δ, for y a + δ,

3 and its Fourier transform satisfies the inequality Diophantine inequality 29 Φ(x = Φ(x min P r o o f. See [4] or [5]. e( xyϕ(y dy ( ( k 2a, π x, k. π x 2π x δ Throughout this paper we denote by ϕ(y the function from Lemma with parameters a = 9 0ε, δ =, k = r, and by Φ(x the Fourier transform of ϕ(y. Lemma 2. Assume that G(x, F (x are real functions defined in [a, b], G(x H for a x b and G(x/F (x is a monotonous function. Set I = b a G(xe(F (x dx. If F (x h > 0 for all x [a, b] or if F (x h < 0 for all x [a, b] then I H/h. Then If F (x h > 0 for all x [a, b] then P r o o f. See [6, p. 7]. I H/ h. Lemma 3. Suppose that f (t exists, is continuous on [a, b] and satisfies f (t λ (λ > 0 for t [a, b]. a<n b P r o o f. See [6, p. 04]. Lemma 4. If 2 t x then P r o o f. See [3, p. 80]. e(f(n (b aλ /2 + λ /2. Ψ(x = x γ t x ϱ ϱ + O ( x log 2 x t.

4 292 D. I. Tolev Lemma 5. Assume that R Y. There exists a number γ > 0 such that if θ(r = log 2/3 (R + 0 log log(r + 0 then the following estimate holds: Y /2 R log 6 Y, Y 3/4 R Y, Y β e 2 log Y +3 log R 2 3 log Y log R log 6 Y, Y /3 R Y 3/4, 0<γ R Y θ(r R 2.4θ(R log 45 Y, R Y /3. P r o o f. This lemma can be deduced from the density theorems of Ingham (see [6, p. 236] and Huxley (see [2] combined with the estimate of I. M. Vinogradov for the zero-free region of ζ(s (see [3, p. 00]. Detailed computations may be found in []. Lemma 6. We have γ I 3 (xe( NxΦ(x dx εx 3 c. P r o o f. Denote the above integral by H. We have H = X/2 X/2 X/2 e((t c + t c 2 + t c 3 NxΦ(x dx dt dt 2 dt 3 (the change of the order of integration is legitimate because of the absolute convergence of the integral. Using the Fourier inversion formula we get H = X/2 X/2 X/2 and by the definition of ϕ(y we get H X/2 X/2 X/2 t c +tc 2 +tc 3 N <4ε/5 ϕ(t c + t c 2 + t c 3 N dt dt 2 dt 3 dt dt 2 dt 3 µx µx λx λx ( M dt 3 dt dt 2, λ and µ are real numbers such that ( /c ( ( (3 2 < < λ < µ < 2 /c c < and M = [X/2, X] [(N 4 5 ε tc t c 2 /c, (N ε tc t c 2 /c ]. Because of (3 and the choice of N, X, ε it is easy to show that in fact M is the right interval of the above intersection. Thus by the mean-value

5 Diophantine inequality 293 theorem ξ t,t 2 H ε µx λx µx λx (ξ t,t 2 /c dt dt 2, X c. Therefore, H εx 3 c, which proves the lemma. (i (ii (iii Lemma 7. We have τ τ n+ uniformly with respect to n. n S 2 (x dx X 2 c log 3 X, I 2 (x dx X 2 c log X, S 2 (x dx X log 3 X P r o o f. We only prove (i. Inequalities (ii and (iii can be proved likewise. We have (4 τ We have U S 2 (x dx = U = X/2<p,p 2 X X/2<p,p 2 X log p log p 2 τ log p log p 2 min Uτ log 2 X + V log 2 X, X/2<n,n 2 X n c nc 2 /τ X/2<n X X/2<n 2 X (n c /τ/c n 2 (n c +/τ/c, V = and by the mean-value theorem X/2<n X X/2<n,n 2 X /τ< n c nc 2 (5 U X + τ X2 c. e((p c p c 2x dx ( τ, p c pc 2 n c nc 2. ( + (n c + /τ /c (n c /τ /c

6 294 D. I. Tolev Obviously V l V l (6 V l = X/2<n,n 2 X l< n c nc 2 2l n c nc 2 and l takes the values 2 k /τ, k = 0,, 2,..., with l X c. We have V l. l X/2<n X X/2<n 2 X (n c +l/c n 2 (n c +2l/c For l /τ and X/2 < n X it is easy to see that Hence (7 V l l (n c + 2l /c (n c + l /c >. X/2<n X ((n c + 2l /c (n c + l /c X 2 c by the mean-value theorem. The conclusion follows from formulas (3 and (4 (7. Define Then Lemma 8. Let a m, b n be arbitrary complex numbers and let τ x K, X /4 < R X /2, W = A = X /4 < L < L 2L, LR X. L<n L X /4 <m R a m b n e((mn c x. W (AB /2 X 3/7 (log X, X /4 <m R a m 2, B = L<n L b n 2. P r o o f. Obviously, we may suppose that τ x K. Take s [, R] whose exact value will be determined later. We define the numbers R i, 0 i Q, in the following way: R 0 = X /4, R i+ = min (R i + s, R, R Q = R. Obviously Q R/s. We have W = L<n L i Q R i <m R i a m b n e((mn c x,

7 Diophantine inequality 295 hence W b n a m e((mn c x L<n L i Q R i <m R i and Cauchy s inequality gives W 2 B a m e((mn c x 2 L<n L i Q R i <m R i BQ a m e((mn c x 2 L<n L i Q R i <m R i = BQ a m a m2 e((m c m c 2n c x. i Q R i <m,m 2 R i L<n L After some rearrangements we obtain ( (8 W 2 BQ a m 2 L<n L i Q R i <m R i + a a m m2 i Q R i <m,m 2 R i m m 2 e((m c m c 2n c x L<n L BQ (AL + a m a m+h h s i Q R i <m R i h f(n = ((m + h c m c n c x. It is easy to see that if L < n L then f (n ((m + h c m c L c 2 x and by Lemma 3 we obtain ((m + h c m c /2 L c/2 x /2 L<n L e(f(n e(f(n, L<n L + ((m + h c m c /2 L c/2 x /2 ((m + h c m c /2 L c/2 K /2 + ((m + h c m c /2 L c/2 τ /2 ((m + h c m c /2 L c/2 K /2 s /2 R (c /2 L c/2 K /2.

8 296 D. I. Tolev We substitute this estimate in (8 to get (9 W 2 BQ(AL + s /2 R (c /2 L c/2 K /2 Σ 0, Clearly Σ 0 Σ 0 = h s X /4 <m R h h s ( h s i Q R i <m R i h X /4 <m R h a m a m+h a m 2 /2( X /4 <m R h a m a m+h. The last estimate, (9 and the inequality Q R/s give us (20 W 2 ABLR(/s + s /2 R (c /2 L (c 2/2 K /2. a m+h 2 /2 sa. We now determine s in such a way that the two summands in brackets are equal. This gives the value s = K /3 R ( c/3 L (2 c/3. The assumption of our lemma and (8 imply easily s R. Substituting s in (20 we get after some calculations W 2 ABX 6/7 (log X 2, i.e. W (AB /2 X 3/7 (log X, which is the desired result. Lemma 9. Suppose that a m, b n are arbitrary complex numbers and that L < L 2L, L X. Set V = a m b n e((mn c x. Then V log X L<n L X /4 <m X/n L<n L X /4 <m X/L a mb ne((mn c x for some complex numbers a m, b n such that a m a m, b n b n. P r o o f. Let q be an odd integer such that 2X/L q 4X/L. We have V = (2 a m b n e((mn c x L<n L X /4 <m X/L X /4 <m X/n q k (q /2 ( k(m m e q

9 a (k m = k (q /2 ( km = a m e q max(, k Diophantine inequality 297 L<n L X /4 <m X/L, b (k max(, k n = b n q X /4 <m X/n a (k m b (k n e((mn c x, ( e km. q Obviously a (k m = a m and because of the well-known estimate e(αk min(b a, / α a<k b we get b (k n b n. The absolute value of the double sum over m and n on the right-hand side of (2 takes its maximum for some value k = k 0. Denote by a m, b n the numbers a (k 0 m, b (k 0 n, multiplied by a sufficiently small positive constant. Obviously this proves the lemma. Lemma 0. Assume that τ x K. Then S(x X 3/4 log 3 X. P r o o f. Without loss of generality we may assume that τ x K. Clearly S(x = V 0 (x V (x + O(X /2, V i (x = Λ(ne(n c x, i = 0,. X /4 <n 2 i X Hence it is sufficient to prove that (22 V 0 (x, V (x X 3/4 log 3 X. Consider, for instance, V 0 (x. We use Vaughan s identity (see [8] to get (23 V 0 (x = S S 2 S 3, S = µ(d log l e((ld c x, d X /4 l X/d S 2 = c k e((kr c x, S 3 = k X /2 r X/k X /4 <m X 3/4 X /4 <n X/m c k log k, a m τ(m. a m Λ(ne((mn c x,

10 298 D. I. Tolev We split the sum S 2 in the following way: (24 S 2 = S ( 2 + S (2 2 + O(X 3/4 log X, S ( 2 = S (2 2 = k X /4 Let us treat S. We have (25 S r X/k c k e((kr c x, X /4 <k X /2 X /4 <r X/k d X /4 l X/d c k e((kr c x. log l e((ld c x. We break the sum over l into sums of the type T L = log l e((ld c x L<l L whose number is O(log X and L < L 2L, L X/d. Abel s transformation formula gives us (26 T L log X max, L 2 [L,L ] L<l L 2 e(g(l g(l = (ld c x. Obviously g (l L c 2 d c x for L l L 2 and by Lemma 3 e(g(l L c/2 d c/2 x /2 + L c/2 d c/2 x /2 L<l L 2 L c/2 d c/2 K /2 + L c/2 d c/2 τ /2 X c/2 K /2 + X c/2 d τ /2 X c/2 K /2. The last estimate, together with (25 and (26, implies (27 S log 2 X X /4+c/2 K /2 X /4. We estimate S ( 2 in the same way to get (28 S ( 2 X/4. To estimate S (2 2 and S 3 we need Lemmas 8 and 9. Consider S 3. We split it in the following way: (29 S 3 = W + W 2 + W 3,

11 W = W 2 = W 3 = Diophantine inequality 299 X /2 <n X 3/4 X /4 <m X/n X /2 <n X 3/4 X /4 <m X/n X /4 <n X /2 We break W into sums of the type W (L = a m Λ(ne((mn c x, Λ(ma n e((mn c x, X /4 <m X /2 a m Λ(ne((mn c x. L<n L X /4 <m X/n a m Λ(ne((mn c x whose number is O(log X and X /2 L < L X 3/4, L 2L. According to Lemma 9 there exist complex numbers a m, b n such that a m a m τ(m, b n Λ(n and W (L = W (L (log X W (L, L<n L X /4 <m X/L Using the well-known inequalities τ 2 (m y log 3 y, m y a mb ne((mn c x. Λ 2 (n y log y and Lemma 8 we obtain ( /2 X W (L L log3 X L log X X 3/7 (log X = X 3/4 log X, hence n y (30 W X 3/4 log 3 X. We estimate W 2 and W 3 in the same way (for W 3, of course, we do not use Lemma 9 to get (3 W 2, W 3 X 3/4 log 3 X. We treat S (2 2 analogously: (32 S (2 2 = U + U 2,

12 300 D. I. Tolev U = U 2 = X /4 <r X /2 X /2 <r X 3/4 X /4 <k X/r c k e((kr c x, X /4 <k X /2 c k e((kr c x. We estimate these sums just as W, W 2, W 3, with the help of Lemmas 8 and 9, using the inequality log 2 k y log 2 y. Thus we get k y (33 U, U 2 X 3/4 log 3 X. The inequality (22 follows from (23, (24, (27 (33. proved. Lemma. If T T then X β Xe (log X T 0<γ T P r o o f. Assume that T X /3. Then using Lemma 5 we have X β X β T 0<γ T 0<γ T /4 0<γ X /3 X β log 45 X X 0.2θ(X/3 Xe (log X/4. If X /3 T X 3/4 then using Lemma 5 we get X β log 6 X e ω, T 0<γ T ω = 2 log X log T 2 3 log X log T. Obviously ω is a convex function of log T. Thus max ω = max(ω T =X /3, ω T X /3 T X 3/4 =X 3/4 = 7 8 log X. Hence log 6 X e ω X 7/8 log 6 X Xe (log X/4. If X 3/4 T T then Lemma 5 gives us X β log 6 X X /2 T T T 0<γ T log 6 X X /2 T /2 Xe (log X/4.. The lemma is

13 Lemma is proved. Lemma 2. If T T then T Diophantine inequality 30 0<γ T X β Xe (log X P r o o f. This lemma follows immediately from Lemma. Lemma 3. If x τ then J(x is defined in (2. J(x Xe (log X/5, P r o o f. Without loss of generality we may assume that x 0. By the definition we have I ϱ (x = X/2 /4 t β e(f (t dt F (t = t c x + γ 2π log t. Define the following sets of non-trivial zeroes of ζ(s: M = {ϱ γ T, γ/2π > 3 2 cxc x}, M 2 = {ϱ γ T, 2 c(x/2c x γ/2π 3 2 cxc x}, M 3 = {ϱ γ T, γ/2π < 2 c(x/2c x}. (The set M 2 may be empty. Let X c x τ. If ϱ M 3 then F (t ( ( X c X 2 and according to Lemma 2 Hence ϱ M 3 I ϱ (x I ϱ (x c x γ > 0 2π X β c(x/2 c x + γ/2π. πc(x/2 c x γ T πc(x/2 c x γ X c x log X max T T X β c(x/2 c x + γ/2π ( X β X c x +. X c x<γ T X β T 0<γ T. X β γ

14 302 D. I. Tolev Using Lemma 2 we obtain (34 I ϱ (x Xe (log X ϱ M 3 If ϱ M then F (t t and by Lemma 2 we get Therefore and Lemma 2 gives (35 /5 ( cx c x + γ ( cx c x + γ < 0 2π X 2π I ϱ (x ϱ M I ϱ (x X β γ/2π cx c x Xβ γ. Xβ γ 0<γ T log X max T T ( ϱ M I ϱ (x Xe (log X. X β T 0<γ T If ϱ M 2 then γ/2π X c x and F (t X c 2 x. According to Lemma 2 Hence, using Lemma we get (36 I ϱ (x Xc ϱ M 2 x I ϱ (x Xβ Xc x. 0<γ X c x /5. X β Xe (log X/4. (If the set M 2 is empty then the last estimate is also true. Inequalities (34 (36 give us (37 J(x Xe (log X/5. If 0 x X c then estimating I ϱ (x trivially we have I ϱ (x X β X β0, ϱ M 2 ϱ M 2 β 0 = max ϱ M2 β <. The terms ϱ M I ϱ (x and ϱ M 3 I ϱ (x can be estimated as in the previous case. Thus (37 is always true. Lemma 4. If x τ then S(x = I(x + O(Xe (log X/5.

15 Diophantine inequality 303 P r o o f. Obviously (38 S(x = U(x + O(X /2, U(x = X/2<n X Using Abel s transformation we get U(x = X/2 Λ(ne(n c x. (Ψ(t Ψ(X/2 d dt (e(tc x dt + (Ψ(X Ψ(X/2e(X c x. According to Lemma 4 we obtain ( U(x = t X 2 t ϱ (X/2 ϱ ϱ X/2 γ T ( X log 2 X d + O T dt (e(tc x dt ( + X X 2 X ϱ (X/2 ϱ ( X log 2 X + O e(x c x ϱ T = + X/2 γ T ( t X 2 ( X X 2 γ T γ T Using partial integration we get U(x = X/2 e(t c x ( t ϱ (X/2 ϱ d ϱ dt (e(tc x dt X ϱ (X/2 ϱ ( τx e(x c +c x + O ϱ T γ T ( τx +c = I(x J(x + O T t ϱ ( τx +c dt + O T log 2 X. log 2 X log 2 X. The conclusion follows from formulas (3, (4, (38 and Lemma 3. P r o o f o f t h e T h e o r e m. It is sufficient to show that the sum B = log p log p 2 log p 3 X/2<p,p 2,p 3 X p c +pc 2 +pc 3 N <ε

16 304 D. I. Tolev tends to infinity as X tends to infinity. Set B = log p log p 2 log p 3 ϕ(p c + p c 2 + p c 3 N. X/2<p,p 2,p 3 X By the definition of ϕ(y we have (39 B B. The Fourier transformation formula gives us B = = X/2<p,p 2,p 3 X log p log p 2 log p 3 S 3 (xe( NxΦ(x dx. Let us represent B in the form (40 B = D + D 2 + D 3, (4 D = D 2 = D 3 = τ S 3 (xe( NxΦ(x dx, τ< x <K x >K Using (5 (8 and Lemma we have D 3 K X 3 S 3 (x Φ(x dx K e((p c + p c 2 + p c 3 NxΦ(x dx S 3 (xe( NxΦ(x dx, S 3 (xe( NxΦ(x dx. ( r ( r r r dx X 3. x 2π x 2π K (42 We next treat D 2 : D 2 τ K S 3 (x Φ(x dx ( max τ x K S(x K τ S 2 (x Φ(x dx.

17 According to Lemma we get τ K S 2 (x Φ(x dx ε ε 0 n /ε Diophantine inequality 305 /ε τ n+ n S 2 (x dx + S 2 (x dx + The last estimate and Lemma 7(iii give τ K K /ε S 2 (x dx x /ε n K S 2 (x Φ(x dx X log 4 X. Therefore, by (42 and by Lemma 0 we have (43 D 2 ε X3 c log X. We now find an asymptotic formula for D. Set H = τ Using Lemmas 7(i, 7(ii and 4 we have (44 Define Then we have D H ε τ τ I 3 (xe( NxΦ(x dx. S 3 (x I 3 (x Φ(x dx S(x I(x ( S 2 (x + I 2 (x dx (log εxe X/5( τ εx 3 c e (log X/6. H = (45 H H S 2 (x dx + I 3 (xe( NxΦ(x dx. τ I 3 (x Φ(x dx. τ n+ S 2 (x dx. n n I 2 (x dx By Lemma 2 we get I(x /( x X c. Therefore, by (45 and Lemma

18 306 D. I. Tolev we obtain (46 H H X 3(c τ Φ(x dx x 3 ε X 3(c τ 2 ε X3 c log X. It now follows from formulas (40, (4, (43 (46 that ( εx 3 c B = H + O. log X Hence by Lemma 6 we have B εx 3 c. Together with (40 this implies that B εx 3 c, and so B as X. The Theorem is proved. Finally, I would like to thank Professor A. A. Karatsuba for useful and encouraging discussions. References [] S. A. Gritsenko, On a problem of I. M. Vinogradov, Mat. Zametki 39 (987, (in Russian. [2] M. N. H u x l e y, On the difference between consecutive primes, Invent. Math. 5 (972, [3] A. A. Karatsuba, Principles of Analytic Number Theory, Nauka, Moscow 983 (in Russian. [4] I. I. Piatetski-Shapiro, On a variant of Waring Goldbach s problem, Mat. Sb. 30 (72 ( (952, (in Russian. [5] B. I. Segal, On a theorem analogous to Waring s theorem, Dokl. Akad. Nauk SSSR (N. S. 2 (933, (in Russian. [6] E. G. Titchmarsh, The Theory of the Riemann Zeta-function (revised by D. R. Heath-Brown, Clarendon Press, Oxford 986. [7] D. I. T o l e v, Diophantine inequalities involving prime numbers, Thesis, Moscow University, 990 (in Russian. [8] R. C. Vaughan, On the distribution of αp modulo, Mathematika 24 (977, DEPARTMENT OF MATHEMATICS UNIVERSITY OF PLOVDIV TSAR ASEN 24 PLOVDIV 4000, BULGARIA Received on (240