# Note on the Riemann Hypothesis

The Riemann Hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1/2. In 1984, Guy Robin stated a new criterion for the Riemann Hypothesis. We prove that the Riemann Hypothesis is true using the Robin's criterion. This was an accepted and exposed presentation at the International Conference on Recent Developments in Mathematics (ICRDM 2022). The Riemann hypothesis belongs to the Hilbert's eighth problem on Hilbert's list of twenty-three unsolved problems. This is one of the Clay Mathematics Institute's Millennium Prize Problems. Practical uses of the Riemann hypothesis include many propositions that are considered to be true under the assumption of the Riemann hypothesis and some of them that can be shown to be equivalent to the Riemann hypothesis. Indeed, the Riemann hypothesis is closely related to various mathematical topics such as the distribution of primes, the growth of arithmetic functions, the Lindelof hypothesis, the Large Prime Gap Conjecture, etc. In general, a proof of the Riemann hypothesis could spur considerable advances in many mathematical areas.

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### Note on the Riemann Hypothesis

• 1. Note on the Riemann Hypothesis International Conference on Recent Developments in Mathematics (ICRDM 2022) Frank Vega, CopSonic France vega.frank@gmail.com August 24-26, 2022 Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 1 / 20
• 2. Abstract The Riemann Hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1 2 . In 1984, Guy Robin stated a new criterion for the Riemann Hypothesis. We prove that the Riemann Hypothesis is true using the Robin’s criterion. Note This result is an extension of my article “Robin’s criterion on divisibility” published by The Ramanujan Journal (03 May 2022). References This presentation and references can be found at my ResearchGate Project: https://www.researchgate.net/project/The-Riemann-Hypothesis. Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 2 / 20
• 3. Introduction The Riemann Hypothesis is considered by many to be the most important unsolved problem in pure mathematics. It was proposed by Bernhard Riemann (1859). The Riemann Hypothesis belongs to the Hilbert’s eighth problem on David Hilbert’s list of twenty-three unsolved problems. This is one of the Clay Mathematics Institute’s Millennium Prize Problems. Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 3 / 20
• 4. Objectives As usual σ(n) is the sum-of-divisors function of n X d|n d, where d | n means the integer d divides n. Define f (n) as σ(n) n . We provide a proof for the Riemann Hypothesis using the properties of the f function. Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 4 / 20
• 5. Methods We say that Robin(n) holds provided that f (n) < eγ · log log n, where the constant γ ≈ 0.57721 is the Euler-Mascheroni constant and log is the natural logarithm. Proposition 1 Robin(n) holds for all natural numbers n > 5040 if and only if the Riemann Hypothesis is true (Robin, 1984, Theorem 1 pp. 188). Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 5 / 20
• 6. Superabundant Numbers A natural number n is called superabundant precisely when, for all natural numbers m < n f (m) < f (n). Let q1 = 2, q2 = 3, . . . , qk denote the first k consecutive primes, then an integer of the form Qk i=1 qai i with a1 ≥ a2 ≥ . . . ≥ ak ≥ 1 is called a Hardy-Ramanujan integer (Choie et al., 2007, pp. 367). If n is superabundant, then n is a Hardy-Ramanujan integer (Alaoglu and Erdős, 1944, Theorem 1 pp. 450). Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 6 / 20
• 7. Central Lemma The following is a key Lemma. Lemma 2 If the Riemann Hypothesis is false, then there are infinitely many superabundant numbers n such that Robin(n) fails (i.e. Robin(n) does not hold). Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 7 / 20
• 8. If the Riemann Hypothesis is false, then there are infinitely many colossally abundant numbers n > 5040 such that Robin(n) fails (Robin, 1984, Proposition pp. 204). Every colossally abundant number is superabundant (Alaoglu and Erdős, 1944, pp. 455). In this way, the proof is complete. ■ Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 8 / 20
• 9. A Strictly Decreasing Sequence For every prime number qk > 2, we define the sequence: Yk = e 0.2 log2(qk ) (1 − 1 log(qk ) ) . As the prime number qk increases, the sequence Yk is strictly decreasing (Vega, 2022, Lemma 6.1 pp. 6). Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 9 / 20
• 10. Known Results We use the following Propositions: Proposition 3 (Vega, 2022, Theorem 6.6 pp. 8). Let Qk i=1 qai i be the representation of a superabundant number n > 5040 as the product of the first k consecutive primes q1 < . . . < qk with the natural numbers a1 ≥ a2 ≥ . . . ≥ ak ≥ 1 as exponents. Suppose that Robin(n) fails. Then, αn < log log(Nk )Yk log log n , where Nk = Qk i=1 qi is the primorial number of order k and αn = Qk i=1 q ai +1 i q ai +1 i −1 . Proposition 4 (Nazardonyavi and Yakubovich, 2013, Lemma 3.3 pp. 8). Let x ≥ 11. For y x, we have log log y log log x r y x . Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 10 / 20
• 11. Main Insight This is the main insight. Lemma 5 Let Qk i=1 qai i be the representation of a superabundant number n 5040 as the product of the first k consecutive primes q1 . . . qk with the natural numbers a1 ≥ a2 ≥ . . . ≥ ak ≥ 1 as exponents. Suppose that Robin(n) fails. Then, αn r (Nk )Yk n , where Nk = Qk i=1 qi is the primorial number of order k and αn = Qk i=1 q ai +1 i q ai +1 i −1 . Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 11 / 20
• 12. When n 5040 is a superabundant number and Robin(n) fails, then we have αn log log(Nk )Yk log log n . We assume that (Nk )Yk n 5040 11 since αn 1. Consequently, log log(Nk )Yk log log n r (Nk )Yk n . As result, we obtain that αn r (Nk )Yk n and thus, the proof is done. ■ Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 12 / 20
• 13. Main Theorem This is the main theorem. Theorem 6 The Riemann Hypothesis is true. Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 13 / 20
• 14. We know there are infinitely many superabundant numbers (Alaoglu and Erdős, 1944, Theorem 9 pp. 454). In number theory, the p-adic order of an integer n is the exponent of the highest power of the prime number p that divides n. It is denoted νp(n). For every prime q, νq(n) goes to infinity as long as n goes to infinity when n is superabundant (Nazardonyavi and Yakubovich, 2013, Theorem 4.4 pp. 12), (Alaoglu and Erdős, 1944, Theorem 7 pp. 454). Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 14 / 20
• 15. Let nk 5040 be a large enough superabundant number such that qk is the largest prime factor of nk . Suppose that Robin(nk ) fails. In the same way, let nk′ be another superabundant number much greater than nk such that Robin(nk′ ) fails too. Under our assumption, we have αnk r (Nk )Yk nk and αnk′ s (Nk′ )Yk′ nk′ . Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 15 / 20
• 16. Hence, αnk′ · αnk s (Nk′ )Yk′ nk′ · αnk . Consequently, αnk′ · αnk s (Nk′ )Yk′ nk′ · r (Nk )Yk nk . So, (αnk′ · αnk )2 (Nk′ )Yk′ nk′ · (Nk )Yk nk . Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 16 / 20
• 17. However, we know that (αnk′ · αnk )2 1. Moreover, we can see that (Nk′ )Yk′ nk′ · (Nk )Yk nk ≤ 1 since the following inequality Yk ≤ log(nk′ · nk ) log((Nk′ ) Y k′ Yk · Nk ) is satisfied for nk′ much greater than nk , because of Yk′ Yk 1 and limk→∞ Yk = 1. In this way, we obtain the contradiction 1 1 under the assumption that Robin(nk ) fails. Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 17 / 20
• 18. To sum up, the study of this arbitrary large enough superabundant number nk 5040 reveals that Robin(nk ) holds on anyway. Accordingly, Robin(n) holds for all large enough superabundant numbers n. This contradicts the fact that there are infinite superabundant numbers n, such that Robin(n) fails when the Riemann Hypothesis is false. By reductio ad absurdum, we prove that the Riemann Hypothesis is true. ■ Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 18 / 20
• 19. Conclusions The Riemann Hypothesis is closely related to various mathematical topics such as the distribution of primes, the growth of arithmetic functions, the Large Prime Gap Conjecture, etc. Certainly, a proof of the Riemann Hypothesis could spur considerable advances in many mathematical areas, such as number theory and pure mathematics in general. Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 19 / 20
• 20. References Leonidas Alaoglu and Paul Erdős. On highly composite and similar numbers. Transactions of the American Mathematical Society, 56(3):448–469, 1944. URL https://doi.org/10.2307/1990319. YoungJu Choie, Nicolas Lichiardopol, Pieter Moree, and Patrick Solé. On Robin’s criterion for the Riemann hypothesis. Journal de Théorie des Nombres de Bordeaux, 19 (2):357–372, 2007. URL https://doi.org/10.5802/jtnb.591. Sadegh Nazardonyavi and Semyon Yakubovich. Superabundant numbers, their subsequences and the Riemann hypothesis. arXiv preprint arXiv:1211.2147v3, 2013. version 3 (Submitted on 26 Feb 2013). arXiv URL https://arxiv.org/pdf/1211.2147v3.pdf. Guy Robin. Grandes valeurs de la fonction somme des diviseurs et hypothèse de Riemann. J. Math. pures appl, 63(2):187–213, 1984. URL http://zakuski.utsa.edu/~jagy/Robin_1984.pdf. Frank Vega. Robin’s criterion on divisibility. The Ramanujan Journal, pages 1–11, 2022. URL https://doi.org/10.1007/s11139-022-00574-4. Frank Vega, CopSonic France (vega.frank@gmail.com) Note on the Riemann Hypothesis August 24-26, 2022 20 / 20
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