The Complete Proof of the Riemann Hypothesis

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 n > 5040, where σ ( n ) is the sum-of-divisors function and γ ≈ 0 . 57721 is the Euler-Mascheroni constant. We show there is a contradiction just assuming the possible smallest counterexample n > 5040 of the Robin inequality. In this way, we prove that the Robin inequality is true for all n > 5040 and thus, the Riemann Hypothesis is true.


Introduction
In mathematics, 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]. As usual σ(n) is the sum-of-divisors function of n [2]: d|n d where d | n means the integer d divides to n and d n means the integer d does not divide to n. Define f (n) to be σ(n) n . Say Robins(n) holds provided f (n) < e γ × log log n.
The constant γ ≈ 0.57721 is the Euler-Mascheroni constant, and log is the natural logarithm. The importance of this property is: holds for all n > 5040 if and only if the Riemann Hypothesis is true [1].
Let q 1 = 2, q 2 = 3, . . . , q m denote the first m consecutive primes, then an integer of the form m i=1 q e i i with e 1 ≥ e 2 ≥ · · · ≥ e m is called an Hardy-Ramanujan integer [2]. A natural number n is called superabundant precisely when, for all m < n f (m) < f (n). Theorem 1.2. If n is superabundant, then n is an Hardy-Ramanujan integer [3].
The smallest counterexample of the Robin inequality greater than 5040 must be a superabundant number [4].
We prove the nonexistence of such counterexample and therefore, the Riemann Hypothesis is true.

Proof of Main Theorems
Let n = s i=1 q e i i be a factorisation of n, where we ordered the primes q i in such a way that e 1 ≥ e 2 ≥ · · · ≥ e s . We say that e = (e 1 , . . . , e s ) is the exponent pattern of the integer n [2]. Note that s i=1 p e i i is the minimal number having exponent pattern e when p 1 = 2, p 2 = 3, . . . , p s denote the first s consecutive primes and e 1 ≥ e 2 ≥ · · · ≥ e s . We denote this (Hardy-Ramanujan) number by m(e) [2].
Theorem 2.1. Let m i=1 q e i i be the representation of n as a product of the primes q 1 < · · · < q m with natural numbers as exponents e 1 , . . . , e m . We obtain a contradiction just assuming that n > 5040 is the smallest integer such that Robins(n) does not hold.
Proof. According to the theorems 1.2 and 1.3, the primes q 1 < · · · < q m must be the first m consecutive primes and e 1 ≥ e 2 ≥ · · · ≥ e m since n > 5040 should be an Hardy-Ramanujan integer. Let e denote the factorisation pattern of n × q m . Based on the result of the article [5], the value n × q m cannot be a square full number [2]. Therefore n × q m > m(e) and consequently, n > m(e) q m . Thus, we have that Robins( m(e) q m ) holds, because of n > 5040 is the smallest integer such that Robins(n) does not hold. We know that f (p e ) > f (q e ) if p < q [2]. In this way, we would have that f ( m(e) q m ) > f (n) since f (q 2 i ) > f (q i ) × f (q m ) for some positive integer 1 ≤ i < m. Certainly, we have that Let's define ω(n) as the number of distinct prime factors of n [2]. From the article [5], we know that ω(n) ≥ 969672728 and the number of primes lesser than q m which have the exponent equal to 1 in n is approximately In this way, there exists a positive integer 1 ≤ i < m such that where we could have that q 2 i n, q i | n, q i+900000000 | n and q 2 i | m(e) q m . Finally, we have that ) < e γ × log log m(e) q m < e γ × log log n.
However, this a contradiction with our initial assumption. To sum up, we obtain a contradiction just assuming that n > 5040 is the smallest integer such that Robins(n) does not hold.
Proof. Due to the theorem 2.1, we can assure there is not any natural number n > 5040 such that Robins(n) does not hold.
Theorem 2.3. The Riemann Hypothesis is true.
Proof. This is a direct consequence of theorems 1.1 and 2.2