The Riemann Hypothesis Could Be True

The Riemann hypothesis has been considered to be the most important unsolved problem in pure mathematics. The David Hilbert’s list of 23 unsolved problems contains the Riemann hypothesis. Besides, it is one of the Clay Mathematics Institute’s Millennium Prize Problems. The Robin criterion states that the Riemann hypothesis is true if and only if the inequality σ ( n ) < e γ × n × log log n holds for all natural numbers n > 5040, where σ ( x ) is the sum-of-divisors function and γ ≈ 0 . 57721 is the Euler-Mascheroni constant. The Nicolas criterion states that the Riemann hypothesis is true if and only if the inequality (cid:81) q ≤ q n qq − 1 > e γ × log θ ( q n ) is satisfied for all primes q n > 2, where θ ( x ) is the Chebyshev function. Using both inequalities, we show that the Riemann hypothesis could be true.


Introduction
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 [1,2]. In mathematics, the Chebyshev function θ(x) is given by where q ≤ x means all the prime numbers q that are less than or equal to x. Let N n = 2 × 3 × 5 × 7 × 11 × · · · × p n denotes a primorial number of order n such that p n is the n th prime number. Thus, θ(q n ) = log N n . Say Nicolas(q n ) holds provided q≤q n q q − 1 > e γ × log θ(q n ).
The constant γ ≈ 0.57721 is the Euler-Mascheroni constant and log is the natural logarithm. The importance of this inequality is: Theorem 1.1. Nicolas(q n ) holds for all prime numbers q n > 2 if and only if the Riemann hypothesis is true [1].
As usual σ(n) is the sum-of-divisors function of n [3]: where d | n means the integer d divides n and d ∤ n signifies that the integer d does not divide n.
Define f (n) to be σ(n) n . Say Robins(n) holds provided f (n) < e γ × log log n.
The importance of this inequality is: If the Riemann Hypothesis is false, then there are infinitely many natural numbers n > 5040 such that Robins(n) does not hold [2].
It is known that Robins(n) holds for many classes of numbers n.We recall that an integer n is said to be square free if for every prime divisor q of n we have q 2 ∤ n [3]. Theorem 1.3. Robins(n) holds for all natural numbers n > 5040 that are square free [3].
Let q 1 = 2, q 2 = 3, . . . , q m be the first m consecutive primes, then an integer of the form m i=1 q a i i with a 1 ≥ a 2 ≥ · · · ≥ a m ≥ 0 is called an Hardy-Ramanujan integer [3]. Based on the theorem 1.2, we know this result: Theorem 1.4. If the Riemann Hypothesis is false, then there are infinitely many natural numbers n > 5040 which are an Hardy-Ramanujan integer and Robins(n) does not hold [3].
We define H = γ − B such that B ≈ 0.2614972128 is the Meissel-Mertens constant [4]. For all real numbers x ≥ 2, the function u(x) is defined as follows For all real numbers x > 1, we define: Definition 1.5. We define another function: Putting all together yields the proof that the inequality ϖ(p) > u(p) is satisfied for a prime number p ≥ 3 if and only if Nicolas(p) holds. In this way, we introduce another criterion for the Riemann hypothesis based on the Nicolas criterion and deduce some of its consequences. 2

Known Results
We know from the constant H, the following formula: We know this property for the Chebyshev function: Mertens second theorem states that: We know these properties for the function f (n): Theorem 2.4. [6]. Let m i=1 q a i i be the representation of n as a product of primes q 1 < · · · < q m with natural numbers as exponents a 1 , . . . , a m . Then, Theorem 2.5. [3]. For all natural numbers n > 1: We know this result for the Riemann zeta function: Finally, we know that: Theorem 2.7. [1]. For all real numbers x ≥ 2: .

A Central Theorem
The following is a key theorem. It gives an upper bound on f (n) that holds for all natural numbers n. The bound is too weak to prove Robins(n) directly, but is critical because it holds for all natural numbers n. Further the bound only uses the primes that divide n and not how many times they divide n. 3 Theorem 3.1. Let n > 1 and let all its prime divisors be q 1 < · · · < q m . Then, Proof. We use that theorem 2.5: Now, for every prime q > 1, Then by theorem 2.6, Putting this together yields the proof:

A Simple Case
We can easily prove that Robins(n) is true for certain kind of numbers: Theorem 4.1. Robins(n) holds for all natural numbers n > 5040 when q ≤ 5, where q is the largest prime divisor of n.
Proof. Let n > 5040 and let all its prime divisors be q 1 < · · · < q m ≤ 5, then we need to prove For all natural numbers n > 5040, we note that e γ × log log(5040) < e γ × log log n and therefore, the proof is complete when q 1 < · · · < q m ≤ 5.

The Function ϖ(x)
Theorem 5.1. The inequality ϖ(p) > u(p) is satisfied for a prime number p ≥ 3 if and only if Nicolas(p) holds.
Proof. We start from the inequality: which is equivalent to We add the following formula to the both sides of the inequality, and due to the theorem 2.1, we obtain that We distribute it and remove B from the both sides: If we apply the exponentiation to the both sides of the inequality, then we have that which means that Nicolas(p) holds. The same happens in the reverse implication. Proof. This is a direct consequence of theorems 1.1 and 5.1.
Proof. We know that lim x→∞ ϖ(x) = 0 for the limits lim x→∞ δ(x) = 0 and lim x→∞ θ(x) x = 1. In this way, this is a consequence from the theorems 2.2 and 2.3.

Robin and Nicolas Inequalities
Theorem 6.1. Let m i=1 q a i i be the representation of an Hardy-Ramanujan integer n > 5040 as a product of primes q 1 < · · · < q m with natural numbers as exponents a 1 ≥ a 2 ≥ · · · ≥ a m ≥ 0. If Robins(n) does not hold, then Nicolas(q m ) holds indeed.
Proof. When Robins(n) does not hold, then f (n) ≥ e γ × log log n.
We assume that Nicolas(q m ) does not hold as well. Consequently, According to the theorem 2.5, However, this implies that N m > n which is a contradiction since n > 5040 is an Hardy-Ramanujan integer.

Ancillary Theorem
Proof. If we add H to then we obtain that according to the theorems 2.1 and 2.6. Therefore, the proof is done.

Main Insight
The next theorem is a main insight.
Theorem 8.1. Let π 2 6 × log log n ′ ≤ log log n for some natural number n > 5040 such that n ′ is the square free kernel of the natural number n. Then Robins(n) holds.
Proof. Let n ′ be the square free kernel of the natural number n, that is the product of the distinct primes q 1 , . . . , q m . By assumption we have that × log log n ′ ≤ log log n.
By the previous theorem 3.1: Suppose by way of contradiction that Robins(n) fails. Then f (n) ≥ e γ × log log n.
We claim that Since otherwise we would have a contradiction. This shows that This is a contradiction since f (n ′ ) is equal to according to the formula f (x) for the square free numbers [3].

Proof of Main Theorem
Theorem 9.1. The Riemann hypothesis could be true.
Proof. We claim that for every sufficiently large Hardy-Ramanujan integer n > 5040, then Robins(n) could always hold. Let m i=1 q a i i be the representation of a sufficiently large Hardy-Ramanujan integer n > 5040 as a product of primes q 1 < · · · < q m with natural numbers as exponents a 1 ≥ a 2 ≥ · · · ≥ a m ≥ 0. Suppose that Robins(n) does not hold and so, the Riemann hypothesis would be false. Hence, f (n) ≥ e γ × log log n.
We use that theorem 2.4, 8 This is equivalent to where N m is the primorial number of order m. If we apply the logarithm to the both sides of the inequality, then log( log log N m log log n ) + log because of log N m = θ(q m ). Let's multiply by −1 the both sides of the inequality, after adding q≤q m 1 q to the both sides of the inequality. This the same as log( log log n log log N m )−log due to the definition 1.5 and the theorems 2.1, 5.1, 5.3, 6.1 and 7.1, where ε = ϖ(q m )−u(q m ) could be a sufficiently small positive real number that goes to 0 when q m tends to infinity. Actually, q m cannot have an upper bound under our assumption, so the positive value ε gets smaller and smaller as the chosen Hardy-Ramanujan integer n grows. In general, if q m would have an upper bound, then our initial assumption (which is that there would be infinitely many natural numbers n > 5040 which are an Hardy-Ramanujan integer and counterexample of the Robin inequality) fails as a consequence of the theorem 8.1. We know that using the theorem 2.6. It is enough to distribute and remove the value of log( π 2 6 ) from the both sides to show that log( log log n log log N m ) + log However, this could be false for a sufficiently small value of ε, since we know that ε tends to 0 as n grows. In addition, we know that log log n log log N m > 1 due to the theorem 1.3. In conclusion, for every sufficiently large Hardy-Ramanujan integer n > 5040, then Robins(n) could always hold. By contraposition, the Riemann hypothesis could be true, because of the theorems 1.2 and 1.4.