The Complexity of Mathematics

. 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. Many consider it to be the most important unsolved problem in pure mathematics. It is one of the seven Millennium Prize Problems selected by the Clay Mathematics Institute to carry a US 1,000,000 prize for the ﬁrst correct solution. We prove the Riemann hypothesis using the Complexity Theory. Number theory is a branch of pure mathematics devoted primarily to the study of the integers and integer-valued functions. The Goldbach’s conjecture is one of the most important and unsolved problems in number theory. Nowadays, it is one of the open problems of Hilbert and Landau. We show the Goldbach’s conjecture is true using the Complexity Theory as well. An important complexity class is 1NSPACE(S(n)) for some S(n). These mathematical proofs are based on if some unary language belongs to 1NSPACE(S(log n)), then the binary version of that language belongs to 1NSPACE(S(n)) and vice versa. ,


The Riemann Hypothesis
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 . Many consider it to be the most important unsolved problem in pure mathematics [16]. It is of great interest in number theory because it implies results about the distribution of prime numbers [16]. It was proposed by Bernhard Riemann (1859), after whom it is named [16]. In 1915, Ramanujan proved that under the assumption of the Riemann hypothesis, the inequality: d|n d < e γ × n × log log n holds for all sufficiently large n, where γ ≈ 0.57721 is the Euler's constant and d|n means that the natural number d divides n [12]. The largest known value that violates the inequality is n = 5040. In 1984, Guy Robin proved that the inequality is true for all n > 5040 if and only if the Riemann hypothesis is true [12]. Using this inequality, we prove the Riemann hypothesis is true.

The Goldbach's conjecture
The Goldbach's original conjecture, written on 7 June 1742 in a letter to Leonhard Euler, states: "... at least it seems that every number that is greater than 2 is the sum of three primes" [6]. This is known as the ternary Goldbach conjecture. We call a prime as a natural number that is greater than 1 and has exactly two divisors, 1 and the number itself [18]. However, the mathematician Christian Goldbach considered 1 as a prime number. Euler replied in a letter dated 30 June 1742 the following statement: "Every even integer greater than 2 can be written as the sum of two primes" [6]. This is known as the strong Goldbach conjecture.
Using Vinogradov's method, Van der Corput and Estermann showed that almost all even numbers can be written as the sum of two primes (in the sense that the fraction of even numbers which can be so written tends towards 1) [5], [7]. In 1973, Chen showed that every sufficiently large even number can be written as the sum of some prime number and a semi-prime [3]. The strong Goldbach conjecture implies the conjecture that all odd numbers greater than 7 are the sum of three odd primes, which is known today as the weak Goldbach conjecture [6]. In 2012 and 2013, Peruvian mathematician Harald Helfgott published a pair of papers claiming to improve major and minor arc estimates sufficiently to unconditionally prove the weak Goldbach conjecture [10], [11]. In this work, we prove the strong Goldbach's conjecture is true.

Theory and Methods
We use o-notation to denote an upper bound that is not asymptotically tight. We formally define o(g(n)) as the set o(g(n)) = {f (n) : for any positive constant c > 0, there exists a constant n 0 > 0 such that 0 ≤ f (n) < c × g(n) for all n ≥ n 0 }.
For example, 2 × n = o(n 2 ), but 2 × n 2 = o(n 2 ) [4]. In theoretical computer science and formal language theory, a regular language is a formal language that can be expressed using a regular expression [2]. The complexity class that contains all the regular languages is REG. The two-way Turing machines may move their head on the input tape into two-way (left and right directions) while the one-way Turing machines are not allowed to move the head on the input tape to the left [14]. The complexity class 1NSPACE(f (n)) is the set of decision problems that can be solved by a nondeterministic one-way Turing machine M , using space f (n), where n is the length of the input [14].

The Complexity of PRIMES
The checking whether a number is prime can be decided in polynomial time by a deterministic Turing machine [1]. This problem is known as P RIM ES [1].
Proof. If we assume that P RIM ES ∈ 1NSPACE(o(log n)), then the unary version should be regular. Certainly, the standard space translation between the unary and binary languages actually works for nondeterministic machines with small space [8]. This means that if some language belongs to 1NSPACE(S(n)), then the unary version of that language belongs to 1NSPACE(S(log n)) [8]. In this way, when P RIM ES ∈ 1NSPACE(o(log n)), then the unary version should be in 1NSPACE(o(log log n)) and we know that REG = 1NSPACE(o(log log n)) [14], [8]. Since we know that the unary version of P RIM ES is non-regular [13], then we obtain that P RIM ES / ∈ 1NSPACE(S(n)) for all S(n) = o(log n).

The Riemann hypothesis
Definition 1. We define the Robin's language L R as follows: where # is the blank symbol and σ(n) = d|n d [12]. We define the language coL R as where coL R is the complement language of L R .
Theorem 2. If the Riemann hypothesis is true, then the Robin's language L R is non-regular.
Proof. We can easily prove this using the Pumping lemma for regular languages [17].
Definition 2. We define the verification Robin's language L V R as follows: Proof. This is trivially true from the definition of these languages.
Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 4 June 2020 doi:10.20944/preprints202002.0379.v6 Proof. The language L V R cannot be computed in 1NSPACE(S(n)) for some S(n) = o(log n), because of this would imply that the problem P RIM ES belongs to 1NSPACE(S(n)) for some S(n) = o(log n) as well. Certainly if this could be true, then we can find m 2 = e γ × p × log log p and check whether the triple (p, 1, m 2 ) is an element of L V R and thus, we could decide whether p is prime. Indeed, a number p is prime if and only if the sum of its divisors is p + 1 [9]. This could be nondeterministically done on input p just choosing arbitrarily another number m 2 , but instead of putting in the work tapes, then this will put with p and 1 in the output tape just using constant space in one-way. We are able to do this, because of m 2 should be polynomially bounded by the input p. After that, we use the space composition reduction just using the previous output of p, 1 and some integer m 2 into a new nondeterministic Turing machine that would decide whether the instance belongs to L V R in 1NSPACE(S(n)) for some S(n) = o(log n) using (p, 1, m 2 ) as input [15]. Since 1NSPACE(S(n)) for some S(n) = o(log n) is closed under 1NSPACE-reductions with constant space, then the whole computation could be done in 1NSPACE(S(n)) for some S(n) = o(log n). However, this would be a contradiction according to Theorem 1, since the language P RIM ES / ∈ 1NSPACE(S(n)) for all S(n) = o(log n). Consequently, we obtain that L V R / ∈ 1NSPACE(S(n)) for all S(n) = o(log n).

Theorem 4. The Riemann hypothesis is true.
Proof. If the Riemann hypothesis is false, then L R ∈ REG or L R is non-regular and its complement coL R is infinite, since every finite set is regular and REG is also closed under complement [15]. Let's assume the possibility of L R ∈ REG. Nevertheless, this implies that the exponentially more succinct version of L R , that is L V R , should be in 1NSPACE(S(n)) for some S(n) = o(log n), because of REG = 1NSPACE(o(log log n)) and the same algorithm that decides L R within 1NSPACE(o(log log n)) could be easily transformed into a slightly modified algorithm that decides L V R within 1NSPACE(S(n)) for some S(n) = o(log n) [14], [8]. Actually, L R is the unary version of L V R due to Lemma 1. As we mentioned before, the standard space translation between the unary and binary languages actually works for nondeterministic machines with small space [8]. This means that if some unary language belongs to 1NSPACE(S(log n)), then the binary version of that language belongs to 1NSPACE(S(n)) [8]. In this way, we obtain that L R / ∈ REG, since it is not possible that L R ∈ 1NSPACE(o(log log n)) under the result of L V R / ∈ 1NSPACE(S(n)) for all S(n) = o(log n) as a consequence of Theorem 3. Consequently, we obtain a contradiction just assuming that the Riemann hypothesis is false and L R ∈ REG. Hence, we obtain that the Riemann hypothesis is true or the Robin's inequality has an infinite number of counterexamples. However, the asymptotic growth rate of the sigma function can be expressed by [12]: where lim sup is the limit superior and σ(n) = d|n d. In this way, if the Robin's inequality has an infinite number of counterexamples, then the previous limit superior should be false. Since this is a previous checked result, then we have the Riemann hypothesis is true as the remaining only option.

The Goldbach's conjecture
Definition 3. We define the Goldbach's language L G as follows: where # is the blank symbol. We define the language coL G as there are not odd primes p and q such that 2 × n = p + q} where coL G is the complement language of L G .
Theorem 5. If the strong Goldbach's conjecture is true, then the Goldbach's language L G is non-regular.
Proof. We can easily prove this using the Pumping lemma for regular languages [17].
Definition 4. We define the verification Goldbach's language L V G as follows: Lemma 2. The Goldbach's language L G is the unary representation of the verification Goldbach's language L V G .
Proof. This is trivially true from the definition of these languages.
Proof. The language L V G cannot be computed in 1NSPACE(S(n)) for some S(n) = o(log n), because of this would imply that the problem P RIM ES belongs to 1NSPACE(S(n)) for some S(n) = o(log n) as well. Certainly, if this could be true, then we can take any number p and check whether p is prime. This could be nondeterministically done on input p just deterministically generating the numbers p + 3 and 3 and nondeterministically choosing an arbitrary number q, but instead of putting in the work tapes, then we will put them to the output tape just using constant space in one-way. After that, we use the space composition reduction just using the previous output of (p + 3, 3, q) as input into a new nondeterministic Turing machine that would decide whether the instance belongs to L V G in 1NSPACE(S(n)) for some S(n) = o(log n). Indeed, the nondeterministic one-way computation will accept this input if and only if the nondeterministic generated number q is equal to p and p is prime. In Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 4 June 2020 doi:10.20944/preprints202002.0379.v6 this reduction, we assume the initial string p has a binary representation with the least significant bit in the first position within the input tape from left to right. In this way, it will be possible to deterministically generate p + 3 in oneway using constant space. Since 1NSPACE(S(n)) for some S(n) = o(log n) is closed under 1NSPACE-reductions with constant space, then the whole computation could be done in 1NSPACE(S(n)) for some S(n) = o(log n) . Nevertheless, this would be a contradiction according to Theorem 1, since the language P RIM ES / ∈ 1NSPACE(S(n)) for all S(n) = o(log n). Consequently, we obtain that L V G / ∈ 1NSPACE(S(n)) for all S(n) = o(log n).
Theorem 7. The strong Goldbach's conjecture is true.
Proof. If the strong Goldbach's conjecture is false, then L G ∈ REG or L G is non-regular and its complement coL G is infinite, since every finite set is regular and REG is also closed under complement [15]. Let's assume the possibility of L G ∈ REG. However, this implies that the exponentially more succinct version of L G , that is L V G , should be in 1NSPACE(S(n)) for some S(n) = o(log n), because we would have REG = 1NSPACE(o(log log n)) and the same algorithm that decides L G within the complexity 1NSPACE(o(log log n)) could be easily transformed into a slightly modified algorithm that decides L V G within 1NSPACE(S(n)) for some S(n) = o(log n) [14], [8]. Actually, L G is the unary version of L V G due to Lemma 2. As we mentioned before, the standard space translation between the unary and binary languages actually works for nondeterministic machines with small space [8]. This means that if some unary language belongs to 1NSPACE(S(log n)), then the binary version of that language belongs to 1NSPACE(S(n)) [8]. Consequently, we obtain that L G / ∈ REG, since it is not possible that L G ∈ 1NSPACE(o(log log n)) under the result of L V G / ∈ 1NSPACE(S(n)) for all S(n) = o(log n) as result of Theorem 6. In this way, we obtain a contradiction just assuming that the strong Goldbach's conjecture is false and L G ∈ REG. In contraposition, we have the strong Goldbach's conjecture is true or this has an infinite number of counterexamples. Since the fraction of even numbers which can be written as the sum of two primes tends towards 1, then the case of infinite number of counterexamples is not possible [5], [7]. Hence, we prove the strong Goldbach's conjecture is true as the remaining only option.