DISPROOFS OF RIEMANN S HYPOTHESIS

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1 In press at Algebras, Groups and Geometreis, Vol. 1, 004 DISPROOFS OF RIEMANN S HYPOTHESIS Chun-Xuan, Jiang P.O.Box 394, Beijing , China and Institute for Basic Research P.O.Box 1577, Palm Harbor, FL 3468, U.S.A. liukxi@public3.bta.net.cn Abstract As it is well known, the Riemann hypothesis on the zeros of the ζ(s) function has been assumed to be true in various basic developments of the 0-th century mathematics, although it has never been proved to be correct. The need for a resolution of this open historical problem has been voiced by several distinguished mathematicians. By using preceding works, in this paper we present comprehensive disproofs of the Riemann hypothesis. Moreover, in 1994 the author discovered the arithmetic function J n (ω) that can replace Riemann s ζ(s) function in view of its proved features: if J n (ω) 0, then the function has infinitely many prime solutions; and if J n (ω) = 0, then the function has finitely many prime solutions. By using the Jiang J (ω) function we prove the twin prime theorem, Goldbach s theorem and the prime theorem of the form x + 1. Due to the importance of resolving the historical open nature of the Riemann hypothesis, comments by interested colleagues are here solicited. AMS mathematics subject classification: Primary 11M6.

2 1. Introduction In 1859 Riemann defined the zeta function[1] ζ(s) = p (1 p s ) 1 = n=1 1 n s, (1) where s = σ + ti, i = 1, σ and t are real, p ranges over all primes. ζ(s) satisfies the functional equation [] From () we have π s s (1 s) Γ( )ζ(s) = π Γ( 1 s )ζ(1 s). () ζ(ti) 0. (3) Riemann conjectured that ζ(s) has infinitely many zeros in 0 σ 1, called the critical strip. Riemann further made the remarkable conjecture that the zeros of ζ(s) in the critical strip all lie on the central line σ = 1/, a conjecture called the famous Riemann hypothesis (RH). It was stated by Hardy in 1914 that infinitely many zeros lie on the line; A. Selberg stated in 194 that a positive proportion at least of all the zeros lie on the line; Levinson stated in 1974 that more than one third of the zeros lie on the line; Conrey stated in 1989 that at least two fifths of the zeros lie on the line. The use of the RH then lead to many mathematical problems, such as the generalized Riemann conjecture, Artin s conjecture, Weil s conjecture, Langlands program, quantum chaos, the hypothetical Riemann flow [3, 4], the zeta functions and L-functions of an algebraic variety and other studies. Similarly, it is possible to prove many theorems by using the RH. However, the RH remains a basically unproved conjecture to this day. In fact, Hilbert properly stated in 199 that the problem of proving or disproving the RH is one of the most important problems confronting 0th century mathematicians. In 000 Griffiths and Graham pointed out that the RH is the first challenging problem for the 1st century. The proof of the RH then become the millennium prize problem. In 1997 we studied the tables of the Riemann zeta function [5] and reached preliminary results indicating that the RH is false [6, 7, 8]. In

3 this paper we present a comprehensive disproof of the RH and show that the computation of all zeros of the ζ(1/ + ti) function done during the past 100 years is in error, as preliminarily indicated in Refs. [9, 10, 11]. Since the RH is false, all theorems and conjectures based to the same are also false.. Disproofs of Riemann s Hypothesis Theorem 1. ζ(s) has no zeros in the critical strip, that is ζ(s) 0, where 0 σ 1. Proof 1. From (1) we have 1 ζ(s) = p (1 1 p s ) = Reθi, (4) where R = p cos(t log p) R p, R p = 1 + 1, (5) p σ pσ θ = θ p, θ p = tan 1 sin(t log p) p p σ cos(t log p). (6) If σ = 0, from (5) we have R p = 1 cos(t log p). If cos(t log p) = 1, we have R p = 0 then R = 0. If σ > 0 from (5) we have R p 0. ζ(s) = 0 if and only if Re ζ(s) = 0 and Im ζ(s) = 0, that is R =. From (5) we have that if cos(t log p) 0 then R p > 1 and if cos(t log p) > 0 then R p < 1. cos(t log p) is independent of the real part σ, but may well depend on primes p and imaginary part t. We write m + (t) for the number of primes p satisfying cos(t log p) > 0, m (t) for the number of primes p satisfying cos(t log p) 0. For cos(t log p) > 0, we have 1 > R p (1 + ti) > R p (0.5 + ti). (7) If m + (t 1 ) is much greater than m (t 1 ) such that R(0.5 + t 1 i) = min. From (5), (6) and (7) we have for given t 1 minr(σ 1 + t 1 i) > minr(1 + t 1 i) > minr(0.5 + t 1 i) > minr(σ + t 1 i) 0, 3

4 (8) θ(σ 1 + t 1 i) = θ(1 + t 1 i) = θ(0.5 + t 1 i) = θ(σ + t 1 i) = const (9) where σ 1 > 1 and 0 σ < 0.5. Since ζ(s) = 1 from (8) we have R max ζ(σ 1 + t 1 i) < max ζ(1 + t 1 i) < max ζ(0.5 + t 1 i) < max ζ(σ + t 1 i). (10) For cos(t log p) < 0 we have 1 < R p (0.5 + ti) < R p (0.4 + ti) < R p (0.3 + ti). (11) If m (t 1 ) is much greater than m + (t 1 ) such that R(0.5 + t 1 i) = max. From (5), (6) and (11) we have for given t 1 maxr(σ 1 + t 1 i) < maxr(0.5 + t 1 i) < maxr(0.4 + t 1 i), < maxr(0.3 + t 1 i) < maxr(σ + t 1 i), (1) θ(σ 1 + t 1 i) = θ(0.5 + t 1 i) = θ(0.4 + t 1 i) where σ 1 > 0.5 and 0 σ < 0.3. Since ζ(s) = 1 from (1) we have R = θ(0.3 + t 1 i) = θ(σ + t 1 i) = const, min ζ(σ 1 + t 1 i) > min ζ(0.5 + t 1 i) > min ζ(0.4 + t 1 i) > (13) min ζ(0.3 + t 1 i) > min ζ(σ + t 1 i) 0. (14) Proof. We define the beta function β(s) = p (1 + p s ) 1 = n=1 λ(n) n s, (15) where λ(1) = 1, λ(n) = ( 1) a 1+ +a k if n = p a 1 1 p a k k, t 0. 4

5 From (15) we have 1 β(s) = p (1 + 1 p s ) = Re θi, (16) where R = p θ = p For cos(t log p) < 0, we have R p, R cos(t log p) p = , (17) p σ pσ θ p, θ p = tan 1 sin(t log p) p σ + cos(t log p). (18) 1 > R p (1 + ti) > R p (0.5 + ti). (19) If m (t 1 ) is much greater than m + (t 1 ) such that R(0.5 + t 1 i) = min. From (17), (18) and (19) we have for given t 1 min R(σ 1 + t 1 i) > min R(1 + t 1 i) > min R(0.5 + t 1 i) > min R(σ + t 1 i) 0, (0) θ(σ 1 + t 1 i) = θ(1 + t 1 i) = θ(0.5 + t 1 i) = θ(σ + t 1 i) = const, (1) where σ 1 > and 0 σ < 0.5. Since β(s) = 1 R from (0) we have max β(σ 1 + t 1 i) < max β(1 + t 1 i) < max β(0.5 + t 1 i) < max β(σ + t 1 i). () For cos(t log p) > 0 we have 1 < R p (0.5 + ti) < R p (0.4 + ti) < R p (0.3 + ti). (3) If m + (t 1 ) is much greater than m (t 1 ) such that that R(0.5+t 1 i) = max. From (17), (18) and (3) we have for given t 1 max R(σ 1 + t 1 i) < max R(0.5 + t 1 i) < max R(0.4 + t 1 i) < max R(0.3 + t 1 i) < max R(σ + t 1 i), (4) 5

6 θ(σ 1 + t 1 i) = θ(0.5 + t 1 i) = θ(0.4 + t 1 i) = θ(0.3 + t 1 i) = θ(σ + t 1 i) = const, (5) where σ 1 > 0.5 and 0 σ < 0.3. Since β(s) = 1 R from (4) we have From (1) and (15) we have min β(σ 1 + t 1 i) > min β(0.5 + t 1 i) > min β(0.4 + t 1 i) > min β(0.3 + t 1 i) > min β(σ + t 1 i) 0. (6) ζ(s) = ζ(s)β(s). (7) In 1896 J. Hadamard and de la Vallee Poussin proved independently ζ(1 + ti) 0. From (7) we have From (8) we have ζ(1 + ti) = ζ( 1 + ti) β(1 + ti) 0. (8) ζ( 1 + ti) 0 and β(1 + ti) 0. (9) ζ(s) and β(s) are the dual functions. From () we have β( 1 + ti). (30) Therefore we have ζ( 1 + ti) 0. (31) In the same way we have ζ( 1 + ti) = ζ(1 4 + ti) β(1 + ti) 0. (3) 4 6

7 From (3) we have In the same way we have As n we have ζ( ti) 0 and β(1 + ti) 0. (33) 4 Proof 3. For σ > 1 we have ζ( 1 + ti) 0. (34) n ζ(ti) 0. (35) log ζ(s) = p m 1 p mσ exp( itm log p). (36) m=1 If ζ(s) had a zero at 1 + ti, then log ζ(σ + ti) would tend to as σ tends to 1 from the right. From (36) we have log ζ(s) = p m 1 p mσ cos(tm log p), (37) m=1 with t replaced by 0, t, t,, Ht, it gives H 1 j=0 ( H j ) log ζ(σ + (H j)ti) + 1 ( H H ) where A = H 1 j=0 log ζ(σ) = p ( H j ) m 1 p mσ A 0, (38) m=1 cos ((H j)tm log p) + 1 = H 1 [1 + cos(tm log p)] H 0, ( H H ) (39) H is an even number. 7

8 From (38) we have (ζ(σ)) 1 ( H H ) H 1 j=0 ( H j ζ(σ + (H j)ti) ) 1. (40) Since ζ( 1 + eti) [5], where e = 1,,, H, from (40) we have ζ( 1 + eti) 0 for sufficiently large even number H. Min ζ( 1 + ti) 0 but 0. The computation of all zeros of ζ( 1 + ti) is error, which satisfies the the error RH. From (39) we have cos θ + 4 cos θ + 3 = (1 + cos θ), cos 4θ + 8 cos 3θ + 8 cos θ + 56 cos θ + 35 = 8(1 + cos θ) 4, cos 5θ + 1 cos 5θ + 66 cos 4θ + 0 cos 3θ +495 cos θ + 79 cos θ + 46 = 3(1 + cos θ) The Arithmetic Function J n (ω) Replacing Riemann s Hypothesis In view of the preceding results, the RH has no value for the study of prime distributions. In 1994 the author discover the arithmetic function J n (ω) [1, 13] that is able to take the place of Riemann s zeta-functions and L-functions because of the following properties: J n (ω) 0, then the function has infinitely many prime solutions; and if J n (ω) = 0, then the function has finitely many prime solutions. By using Jiang s J n (ω) function we have proved numerous theorems including the twin prime theorem, Goldbach s theorem, the prime k- tuples theorem, Santilli s theory for a prime table, the theorem of finite Fermat primes, the theorem of finite Mersenne primes, the theorem of finite repunit primes, there are infinitely many triples of n, n+1 and n+ that each is the product of k distinct primes, there are infinitely many Carmichael numbers which are product of exactly five primes, there there are finitely many Carmichael numbers which are product of exactly six primes in the prime distributions [14]. We give some theorems below Theorem. Twin prime theorem: p 1 = p +. 8

9 the arithmetic function J (ω) = (p ) 0. Since J (ω) 0, there are infinitely many primes p such that p 1 is a prime. Theorem 3. Goldbach theorem: N = p 1 + p. the arithmetic function J (ω) = (p ) p N p 1 p 0. Since J (ω) 0, every even number N greater than 4 is the sum of two odd primes. Theorem 4. p 1 = (p + 1) + 1. the arithmetic function J (ω) = (p ( 1) p 1 ) 0. Since J (ω) 0, there are infinitely many primes p such that p 1 is a prime. Theorem 5. p 1 = p. J (ω) = (p ( )) 0. p Since J (ω) 0, there are infinitely many primes p such that p 1 is a prime. Theorem 6. p 1 = p + 4 and p = 4p + 1. J (ω) = 3 (p 3) 0. 7 p p i Since J (ω) 0, there are infinitely many primes p such that p 1 and p are primes. 9

10 Theorem 7. p 1 = (p + 1) + 1 and p = (p + 1) + 3. J (ω) = (p 3 ( 1) p 1 ( 3 )) 0. 5 p p i p Since J (ω) 0, there are infinitely many primes p such that p 1 and p are primes. Theorem 8. The prime 13-tuples theorem: p+b: b = 0, 4, 6, 10, 16, 18, 4, 8, 34, 40, 46, 48, 90. Since J (13) = 0, there are no prime 13-tuples if p 13. Theorem 9. The prime 14-tuples theorem: p+b: b = 0,, 6, 8, 1, 18, 0, 6, 30, 3, 36, 4, 48, 50. J (ω) = 300 (p 14) 0. 9 p p i Since J (ω) 0, there are infinitely many prime 14-tuples. Theorem 10. p 1 = 6m + 1, p = 1m + 1, p 3 = 18m + 1, p 4 = 36m + 1, p 5 = 7m + 1. J (ω) = 1 (p 6) 0. 1 Since J (ω) 0, there are infinitely many integers m such that p 1, p, p 3, p 4 and p 5 are primes. n = p 1 p p 3 p 4 p 5 is the Carmichael numbers. Theorem 11. p 3 = p 1 + p + p 1 p. J 3 (ω) = (p 3p + 3) 0. Since J 3 (ω) 0, there are infinitely many primes p 1 and p such that p 3 is a prime. Theorem 1. p 3 = (p 1 + 1) 5 + p. J 3 (ω) = (p 3p + 3) 0. 10

11 Since J 3 (ω) 0, there are infinitely many primes p 1 and p such that p 3 is a prime. Theorem 13. p 4 = p 1 (p + p 3 ) + p p 3. J 4 (ω) = ( (p 1) 4 ) p Since J 4 (ω) 0, there are infinitely many primes p 1, p and p 3 such that p 4 is a prime. Theorem 14. Each of n, n + 1 and n + is the product of k distinct primes. Suppose that each of m 1, m = m 1 +1 and m 3 = m 1 + is the product of k 1 distinct primes. We define p 1 = m m 3 x + 1, p = m 1 m 3 x + 1, p 3 = m 1 m x + 1. (41) the arithmetic function J (ω) = (p 4 χ(p)) 0, (4) where χ(p) = if p m 1 m m 3 ; χ(p) = 0 otherwise. Since J (ω) 0, there exist infinitely many integers x such that p 1, p and p 3 are primes. From (41) we have n = m 1 p 1 = m 1 m m 3 x+m 1, n+1 = m 1 p 1 +1 = m 1 m m 3 x + m = m (m 1 m 3 x + 1) = m p, n + = m 1 p 1 + = m 1 m m 3 x + m 1 + = m 3 (m 1 m x + 1) = m 3 p 3. If p 1, p and p 3 are primes, then each of n, n + 1 and n + is the product of k distinct primes. For example, n = = , n + 1 = = , n + = = J n (ω) is a generalization of Euler s proof for the existence of infinitely many primes. It has a wide application in various fields. Acknowledgements The Author would like to express his deepest appreciation to Professors R. M. Santilli, G. Weiss, D. Zuckerman, Ke-xi Liu, Mao-xian Zuo, Zhongdian Wang, Chang-pei Wang, and Xin-ping Tong for their helps and supports. 11

12 References [1] B. Riemann, Uber die Anzahl der Primzahlen under einer gegebener Grösse, Monatsber. Akad. Berlin (1859). [] H. Davenport, Multiplicative Number Theory, Springer Verlag (1980). [3] N. Katz and P. Sarnak, Zeroes of zeta functions and symmetry, Bull. AMS, 36, 1-6 (1999). [4] A. Connes, Trace formula in noncommutative geometry and the zeros of the Riemann zeta function, Sel. Math., New Ser. 5, 9-106(1999). [5] C. B. Haslgrove, Tables of the Riemann zeta function, Roy. Soc. Math. Tables, Vol.6, Cambridge Univ. Press, London and New York (1960). [6] Chun-xuan, Jiang, The study of the Riemann Zeta function, Unpublished Oct. (1997). [7] Chun-xuan,Jiang, Foundations of Santilli s isonumber theory II, Algebras Groups and Geometries, 15, (1998). [8] Chun-xuan, Jiang, Foundations of Santilli s isonumber theory. In: Foundamental open problems in sciences at the end of the millennium, T. Gill, K. Liu and E. Trell (Eds) Hadronic Press, USA, (1999). [9] Richard P. Brent, On the zeros of the Riemann zeta function in the critical strip, Math. Comp., 33, (1979). [10] J. van de Lune, H. J. J. te Riele and D. T. Winter, On the zeros of the Riemann zeta function in the critical strip, Math. Comp. 46, (1986). [11] A. M. Odlyzko, th zero of the Riemann zeta function and its neighbors, Preprint (1989). 1

13 [1] Chun-xuan, Jiang, On the Yu-Goldbach prime theorem, Guangxi Sciences (in Chinese) (1996). [13] Chun-xuan, Jiang, Foundations of Santilli s isonumber theory I, [14] C.X.Jiang. Foundation of Santilli s isonumber theory with applications to new cryptograms, Fermat s theorem and Goldbach s conjecture. International Academic Press, 00 (also available in dpf format at 13

Euclid-Euler-Jiang Prime Theorem

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