On Sol´e and Planat criterion for the Riemann Hypothesis

There are several statements equivalent to the famous Riemann hypothesis. In 2011, Sol´e and Planat stated that the Riemann hypothesis is true if and only if the inequality ζ (2) · (cid:81) q ≤ q n (1 + 1 q ) > e γ · log θ ( q n ) holds for all prime numbers q n > 3, where θ ( x ) is the Chebyshev function, γ ≈ 0 . 57721 is the Euler-Mascheroni constant, ζ ( x ) is the Riemann zeta function and log is the natural logarithm. In this note, using Sol´e and Planat criterion, we prove that the Riemann hypothesis is true.


Introduction
The Riemann hypothesis is the assertion that all non-trivial zeros have real part 1 2 . It 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. In mathematics, the Chebyshev function θ(x) is given by with the sum extending over all prime numbers q that are less than or equal to x, where log is the natural logarithm. Let's state a property for this function: where q k is the kth prime number (We also use the notation q n to denote the nth prime number).
In mathematics, Ψ(n) = n · q|n 1 + 1 q is called the Dedekind Ψ function, where q | n means the prime q divides n. We say that Dedekind(q n ) holds provided that Next, we have Solé and Planat Theorem: There are several statements out from the Riemann hypothesis condition.
Proposition 1.5. Unconditionally on Riemann hypothesis, there are infinitely many prime numbers q n such that Dedekind(q n ) holds [7, Theorem 4.1 pp. 5].
The following property is based on natural logarithms: Putting all together yields a proof for the Riemann hypothesis using the Chebyshev function.
2 What if the Riemann hypothesis were false?
Several analogues of the Riemann hypothesis have already been proved. Many authors expect (or at least hope) that it is true. However, there are some implications in case of the Riemann hypothesis might be false.
Lemma 2.1. If the Riemann hypothesis is false, then there are infinitely many prime numbers q n for which Dedekind(q n ) fails (i.e. Dedekind(q n ) does not hold).
Proof. The Riemann hypothesis is false, if there exists some natural number We know the bound [7, Theorem 4.2 pp. 5]: where f was introduced in the Nicolas paper [5, Theorem 3 pp. 376]: When the Riemann hypothesis is false, then there exists a real number b < 1 2 for which there are infinitely many natural numbers x such that log f (x) = Ω + (x −b ) [5, Theorem 3 (c) pp. 376]. According to the Hardy and Littlewood definition, this would mean that for every possible positive value of k when b < 1 2 . In this way, this implies that Hence, if the Riemann hypothesis is false, then there are infinitely many natural numbers x such that log f ( , then it would be infinitely many natural numbers x 0 such that log g(x 0 ) > 0. In addition, if log g(x 0 ) > 0 for some natural number x 0 ≥ 5, then log g(x 0 ) = log g(q n ) where q n is the greatest prime number such that q n ≤ x 0 . Actually, according to the definition of the Chebyshev function.

Central Lemma
Proof. We obtain that by Propositions 1.2 and 1.3.

A New Criterion
Theorem 4.1. Dedekind(q n ) holds if and only if the inequality is satisfied for the prime number q n , where the set S = {x : x > q n } contains all the real numbers greater than q n and χ S is the characteristic function of the set S (This is the function defined by χ S (x) = 1 when x ∈ S and χ S (x) = 0 otherwise).
Proof. When Dedekind(q n ) holds, we apply the logarithm to the both sides of the inequality: after of using the Lemma 3.1. Let's distribute the elements of the previous inequality to obtain that when Dedekind(q n ) holds. The same happens in the reverse implication.

The Main Insight
Theorem 5.1. The Riemann hypothesis is true if the inequality is satisfied for all sufficiently large prime numbers q n .
Proof. For large enough prime q n , if Dedekind(q n+1 ) holds then after subtracting the value of log(1 + 1 q n+1 ) to the both sides. Thus, since log log θ(q n ) − log log θ(q n ) = 0. If we obtain that which means that Dedekind(q n ) holds by Theorem 4.1. Hence, it is enough to guarantee that log log θ(q n+1 ) − log log θ(q n ) − log(1 + 1 q n+1 ) ≥ 0 to assure that Dedekind(q n ) holds for a large enough prime number q n when Dedekind(q n+1 ) holds. Since there are infinitely many prime numbers q n+1 > 5 such that Dedekind(q n+1 ) holds, then we can guarantee that Dedekind(q n ) holds as well when holds for all pairs (q n , q n+1 ) of consecutive large enough primes such that q n < q n+1 , then we can confirm that Dedekind(q n ) always holds for all large enough prime numbers q n by Theorem 4.1. As result, if the inequality log log θ(q n+1 ) − log log θ(q n ) − log(1 + 1 q n+1 ) ≥ 0 is satisfied for all sufficiently large prime numbers q n , then there won't exist infinitely many prime numbers q n such that Dedekind(q n ) fails and so, the Riemann hypothesis must be true by Lemma 2.1. Let's distribute the elements of the previous inequality to obtain that 6 The Main Theorem Theorem 6.1. The Riemann hypothesis is true.
Proof. The Riemann hypothesis is true when is satisfied for all sufficiently large prime numbers q n because of the Theorem 5.1. That is the same as We know that q n+1 · log θ(q n+1 ) θ(q n ) ≥ log θ(q n ).

Conclusions
Practical uses of the Riemann hypothesis include many propositions that are known to be true under the Riemann hypothesis and some 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 Lindelöf hypothesis, 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.