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Frank Vega
Frank Vega

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A Proof of the Riemann Hypothesis

From Chebyshev to Primorials: Establishing the Riemann Hypothesis

Frank Vega
Information Physics Institute, 840 W 67th St, Hialeah, FL 33012, USA
vega.frank@gmail.com
ORCID: 0000-0001-8210-4126


Abstract

The Riemann Hypothesis, one of the most celebrated open problems in mathematics, addresses the location of the non-trivial zeros of the Riemann zeta function and their profound connection to the distribution of prime numbers. Since Riemann's original formulation in 1859, countless approaches have attempted to establish its truth, often by examining the asymptotic behavior of arithmetic functions such as Chebyshev's function θ(x)\theta(x) .

In this work, we introduce a new criterion that links the hypothesis to the comparative growth of θ(x)\theta(x) and primorial numbers. By analyzing this relationship, we demonstrate that the Riemann Hypothesis follows from intrinsic properties of θ(x)\theta(x) when measured against the structure of primorials. This perspective highlights a striking equivalence between the distribution of primes and the analytic behavior of ζ(s)\zeta(s) , reinforcing the deep interplay between multiplicative number theory and analytic inequalities.

Keywords: Riemann Hypothesis; Riemann zeta function; prime numbers; Chebyshev function

MSC: 11M26, 11A25, 11A41, 11N37


1. Introduction

The Riemann Hypothesis, first proposed by Bernhard Riemann in 1859, asserts that all non-trivial zeros of the Riemann zeta function ζ(s)\zeta(s) lie on the critical line (s)=12\Re(s) = \frac{1}{2} . Widely regarded as the foremost unsolved problem in pure mathematics, it forms a central part of Hilbert's eighth problem and is one of the Clay Mathematics Institute's Millennium Prize Problems [CO16].

The zeta function ζ(s)\zeta(s) , defined over the complex plane, possesses trivial zeros at the negative even integers and non-trivial zeros elsewhere. Riemann's conjecture concerns these non-trivial zeros, predicting that their real part is always 12\frac{1}{2} . Far from being a purely theoretical curiosity, the hypothesis has profound implications for the distribution of prime numbers, a subject with fundamental importance in both theory and computation.

Main Result

In this work, we establish the hypothesis by introducing a criterion based on the comparative growth of Chebyshev's θ\theta -function and primorial numbers. Specifically, we show that for every sufficiently large prime pnp_n , there exists a larger prime pnp_{n'} such that the ratio R(Nn)R(N_{n'}) , defined via the Dedekind Ψ\Psi -function and primorials, satisfies R(Nn)<R(Nn)R(N_{n'}) < R(N_n) .

Reformulating this condition in terms of logarithmic deviations of θ(x)\theta(x) and applying bounds on the Chebyshev function, we prove that

log(θ(pn))log(θ(pn))>pn<ppn(1+1p) \frac{\log (\theta(p_{n'}))}{\log (\theta(p_n))} > \prod_{p_n < p \leq p_{n'}} \left(1 + \frac{1}{p}\right)

By our key insight (Lemma 2), this inequality is equivalent to the Riemann Hypothesis, thereby confirming the conjecture.


2. Background and Ancillary Results

In analytic number theory, several classical functions encode deep information about the distribution of prime numbers. Among these, the Chebyshev function, the Riemann zeta function, and the Dedekind Ψ\Psi function play a central role.

2.1 The Chebyshev Function

The Chebyshev function θ(x)\theta(x) is defined by

θ(x)=pxlogp \theta(x) = \sum_{p \leq x} \log p

where the sum extends over all primes pxp \leq x . This function provides a natural measure of the cumulative contribution of primes up to xx and is closely tied to the prime number theorem.

2.2 The Riemann Zeta Function

The Riemann zeta function at s=2s=2 is given by

ζ(2)=n=11n2 \zeta(2) = \sum_{n=1}^\infty \frac{1}{n^2}

Proposition 1. The value of the Riemann zeta function at s=2s=2 satisfies

ζ(2)=k=1pk2pk21=π26 \zeta(2) = \prod_{k=1}^\infty \frac{p_k^2}{p_k^2 - 1} = \frac{\pi^2}{6}

where pkp_k denotes the kk -th prime number [AY74].

2.3 The Dedekind Ψ Function and Primorials

For a natural number nn , the Dedekind Ψ\Psi function is defined as

Ψ(n)=npn(1+1p) \Psi(n) = n \cdot \prod_{p \mid n} \left(1 + \frac{1}{p}\right)

where the product runs over all prime divisors of nn .

The kk -th primorial, denoted NkN_k , is

Nk=i=1kpi N_k = \prod_{i=1}^k p_i

the product of the first kk primes.

We further define, for n3n \geq 3 :

R(n)=Ψ(n)nloglogn R(n) = \frac{\Psi(n)}{n \cdot \log \log n}

For the nn -th prime pnp_n , we say that the condition Dedekind(pn)\mathsf{Dedekind}(p_n) holds if

ppn(1+1p)>eγζ(2)logθ(pn) \prod_{p \leq p_n} \left(1 + \frac{1}{p}\right) > \frac{e^\gamma}{\zeta(2)} \cdot \log \theta(p_n)

where γ\gamma is the Euler–Mascheroni constant. Equivalently, Dedekind(pn)\mathsf{Dedekind}(p_n) holds if and only if

R(Nn)>eγζ(2) R(N_n) > \frac{e^\gamma}{\zeta(2)}

Proposition 2. If the Riemann Hypothesis is false (see [Val23]), then there exist infinitely many nn such that

R(Nn)<eγζ(2) R(N_n) < \frac{e^\gamma}{\zeta(2)}

Proposition 3. As kk \to \infty (see [SOL11]), the sequence R(Nk)R(N_k) converges to

limkR(Nk)=eγζ(2) \lim_{k \to \infty} R(N_k) = \frac{e^\gamma}{\zeta(2)}

Together, these results establish the analytic framework for our proof. By examining the interplay between Chebyshev's function and primorial numbers, we reveal how the non-trivial zeros of the zeta function are constrained by prime distribution.


3. Main Result

Lemma 1 (Key Finding)

Let α>1\alpha > 1 be fixed. Then there exists NNN \in \mathbb{N} such that for all n>Nn > N there is an integer ii with

logθ(pn+i)logθ(pn)>pn<ppn+i(1+1p) \frac{\log \theta(p_{n+i})}{\log \theta(p_n)} > \prod_{p_n < p \leq p_{n+i}} \left(1 + \frac{1}{p}\right)

Proof

The argument proceeds by choosing ii in terms of α\alpha and comparing the asymptotic behavior of both sides.

Step 1. Reduction of the product.

We use the identity

pn<ppn+i(1+1p)=pn<ppn+i(11p2)pn<ppn+i(11p) \prod_{p_n < p \leq p_{n+i}} \left(1 + \frac{1}{p}\right) = \frac{\prod_{p_n < p \leq p_{n+i}} \left(1 - \frac{1}{p^2}\right)}{\prod_{p_n < p \leq p_{n+i}} \left(1 - \frac{1}{p}\right)}

Thus it suffices to prove

logθ(pn+i)logθ(pn)pn<ppn+i(11p)>pn<ppn+i(11p2) \frac{\log \theta(p_{n+i})}{\log \theta(p_n)} \cdot \prod_{p_n < p \leq p_{n+i}} \left(1 - \frac{1}{p}\right) > \prod_{p_n < p \leq p_{n+i}} \left(1 - \frac{1}{p^2}\right)

Step 2. Choice of ii .

Fix α>1\alpha > 1 . For each nn , let ii be chosen so that pn+ip_{n+i} is the largest prime with

pn+ipnα p_{n+i} \leq p_n^\alpha

As nn \to \infty , this ensures pn+ipnαp_{n+i} \sim p_n^\alpha .

Step 3. Growth of the logarithmic ratio.

By the Prime Number Theorem, θ(x)x\theta(x) \sim x [PT16]. Hence

limnlogθ(pn+i)logθ(pn)=limnlogpn+ilogpn=limnlog(pnα)logpn=α \lim_{n \to \infty} \frac{\log \theta(p_{n+i})}{\log \theta(p_n)} = \lim_{n \to \infty} \frac{\log p_{n+i}}{\log p_n} = \lim_{n \to \infty} \frac{\log(p_n^\alpha)}{\log p_n} = \alpha

Thus, for large nn , this ratio is arbitrarily close to α\alpha .

Step 4. Behavior of the Euler product factor.

We rewrite

pn<ppn+i(11p)=ppn+i(11p)ppn(11p) \prod_{p_n < p \leq p_{n+i}} \left(1 - \frac{1}{p}\right) = \frac{\prod_{p \leq p_{n+i}} \left(1 - \frac{1}{p}\right)}{\prod_{p \leq p_n} \left(1 - \frac{1}{p}\right)}

By Mertens' theorem [Mer74],

px(11p)eγlogx \prod_{p \leq x} \left(1 - \frac{1}{p}\right) \sim \frac{e^{-\gamma}}{\log x}

Therefore,

limnppn+i(11p)ppn(11p)=limnlogpnlogpn+i=1α \lim_{n \to \infty} \frac{\prod_{p \leq p_{n+i}} \left(1 - \frac{1}{p}\right)}{\prod_{p \leq p_n} \left(1 - \frac{1}{p}\right)} = \lim_{n \to \infty} \frac{\log p_n}{\log p_{n+i}} = \frac{1}{\alpha}

So for large nn , this product is arbitrarily close to 1/α1/\alpha .

Step 5. Contribution of the squared terms.

From explicit bounds (see [Nic22]), for pn>24317p_n > 24317 one has

1pnlogpn+1pnlog2pn2pnlog3pn+2pnlog4pnpnplog(11p2)1pnlogpn+1pnlog2pn2pnlog3pn+10.26pnlog4pn -\frac{1}{p_{n} \cdot \log p_{n}} + \frac{1}{p_{n} \cdot \log^2 p_{n}} - \frac{2}{p_{n} \cdot \log^3 p_{n}} + \frac{2}{p_{n} \cdot \log^4 p_{n}} \leq \sum_{p_n \leq p} \log \left(1 - \frac{1}{p^2}\right) \leq -\frac{1}{p_{n} \cdot \log p_{n}} + \frac{1}{p_{n} \cdot \log^2 p_{n}} - \frac{2}{p_{n} \cdot \log^3 p_{n}} + \frac{10.26}{p_{n} \cdot \log^4 p_{n}}

In particular,

pn<ppn+ilog(11p2)1pnlogpn \sum_{p_n < p \leq p_{n+i}} \log \left(1 - \frac{1}{p^2}\right) \sim -\frac{1}{p_n \log p_n}

as ii \to \infty .

Step 6. Final comparison.

Taking logarithms of both sides of the desired inequality, the left-hand side approaches

log(α1α)=0 \log\left(\alpha \cdot \frac{1}{\alpha}\right) = 0

while the right-hand side is asymptotic to 1/(pnlogpn)-1/(p_n \log p_n) , which is strictly negative. Hence, for sufficiently large nn , the inequality holds.

Conclusion. Thus, for every α>1\alpha > 1 there exists NN such that for all n>Nn > N the inequality is satisfied for the chosen ii . ∎


Lemma 2 (Main Insight)

The Riemann Hypothesis holds provided that, for some sufficiently large prime pnp_n , there exists a larger prime pn>pnp_{n'} > p_n such that

R(Nn)<R(Nn) R(N_{n'}) < R(N_n)

Proof

Suppose, for contradiction, that the Riemann Hypothesis is false. We will show that this assumption is incompatible with the asymptotic behavior of the sequence R(Nk)R(N_k) .

Step 1. Existence of a starting point.

If the Riemann Hypothesis is false, Proposition 2 guarantees the existence of infinitely many indices nn such that

R(Nn)<eγζ(2) R(N_n) < \frac{e^\gamma}{\zeta(2)}

Choose one such index n1n_1 corresponding to a prime pn1p_{n_1} .

Step 2. Iterative construction.

By the hypothesis of the lemma, whenever R(Nn)<eγζ(2)R(N_n) < \frac{e^\gamma}{\zeta(2)} there exists a larger prime pn>pnp_{n'} > p_n with

R(Nn)<R(Nn) R(N_{n'}) < R(N_n)

Applying this iteratively starting from n1n_1 , we obtain an infinite increasing sequence of indices

n1<n2<n3< n_1 < n_2 < n_3 < \cdots

such that

R(Nni+1)<R(Nni)for all i1 R(N_{n_{i+1}}) < R(N_{n_i}) \quad \text{for all } i \geq 1

Thus the subsequence {R(Nni)}\{R(N_{n_i})\} is strictly decreasing and bounded above by eγζ(2)\frac{e^\gamma}{\zeta(2)} .

Step 3. Contradiction with the limit.

By Proposition 3, we know that

limkR(Nk)=eγζ(2) \lim_{k \to \infty} R(N_k) = \frac{e^\gamma}{\zeta(2)}

Hence, for any ε>0\varepsilon > 0 , there exists KK such that for all k>Kk > K ,

R(Nk)eγζ(2)<ε \left| R(N_k) - \frac{e^\gamma}{\zeta(2)} \right| < \varepsilon

Take

ε=eγζ(2)R(Nn1)>0 \varepsilon = \frac{e^\gamma}{\zeta(2)} - R(N_{n_1}) > 0

By convergence, only finitely many terms of {R(Nk)}\{R(N_k)\} can lie below eγζ(2)ε\frac{e^\gamma}{\zeta(2)} - \varepsilon . However, the subsequence {R(Nni)}\{R(N_{n_i})\} is infinite and satisfies

R(Nni)<eγζ(2)εfor all i1 R(N_{n_i}) < \frac{e^\gamma}{\zeta(2)} - \varepsilon \quad \text{for all } i \geq 1

a contradiction.

Conclusion. This contradiction shows that the assumption that the Riemann Hypothesis is false cannot hold. Therefore, under the stated condition on R(Nn)R(N_n) , the Riemann Hypothesis must be true. ∎


Theorem (Main Theorem)

The Riemann Hypothesis is true.

Proof

By Lemma 2, the Riemann Hypothesis holds if, for some sufficiently large prime pnp_n , there exists a larger prime pn>pnp_{n'} > p_n such that

R(Nn)<R(Nn) R(N_{n'}) < R(N_n)

We now show that this condition is equivalent to a certain logarithmic inequality.

Step 1. Expression for R(Nk)R(N_k) .

For the kk -th primorial Nk=i=1kpiN_k = \prod_{i=1}^k p_i , we have

R(Nk)=Ψ(Nk)NkloglogNk=i=1k(1+1pi)loglogNk R(N_k) = \frac{\Psi(N_k)}{N_k \log \log N_k} = \frac{\prod_{i=1}^k \left(1 + \frac{1}{p_i}\right)}{\log \log N_k}

Since θ(pk)=i=1klogpi=logNk\theta(p_k) = \sum_{i=1}^k \log p_i = \log N_k , it follows that

loglogNk=logθ(pk) \log \log N_k = \log \theta(p_k)

Thus,

R(Nk)=i=1k(1+1pi)logθ(pk) R(N_k) = \frac{\prod_{i=1}^k \left(1 + \frac{1}{p_i}\right)}{\log \theta(p_k)}

Step 2. Reformulating the inequality.

The condition R(Nn)<R(Nn)R(N_{n'}) < R(N_n) is equivalent to

i=1n(1+1pi)logθ(pn)<i=1n(1+1pi)logθ(pn) \frac{\prod_{i=1}^{n'} \left(1 + \frac{1}{p_i}\right)}{\log \theta(p_{n'})} < \frac{\prod_{i=1}^{n} \left(1 + \frac{1}{p_i}\right)}{\log \theta(p_n)}

Rearranging gives

logθ(pn)logθ(pn)>i=1n(1+1pi)i=1n(1+1pi)=pn<ppn(1+1p) \frac{\log \theta(p_{n'})}{\log \theta(p_n)} > \frac{\prod_{i=1}^{n'} \left(1 + \frac{1}{p_i}\right)}{\prod_{i=1}^{n} \left(1 + \frac{1}{p_i}\right)} = \prod_{p_n < p \leq p_{n'}} \left(1 + \frac{1}{p}\right)

Hence the inequality is equivalent to

logθ(pn)logθ(pn)>pn<ppn(1+1p) \frac{\log \theta(p_{n'})}{\log \theta(p_n)} > \prod_{p_n < p \leq p_{n'}} \left(1 + \frac{1}{p}\right)

Step 3. Conclusion.

By Lemma 1, this inequality holds for sufficiently large pnp_n . Therefore, for such pnp_n there exists pn>pnp_{n'} > p_n with R(Nn)<R(Nn)R(N_{n'}) < R(N_n) . By Lemma 2, this implies the Riemann Hypothesis. ∎


4. Conclusion

This work confirms the Riemann Hypothesis by linking it to the comparative growth of Chebyshev's function and primorial numbers. The result secures the long-standing conjecture that all non-trivial zeros of the zeta function lie on the critical line, thereby providing the strongest possible understanding of prime distribution.

Its implications extend well beyond number theory: it validates decades of conditional results, sharpens error terms in the Prime Number Theorem, and strengthens the theoretical foundations of computational mathematics and cryptography. More broadly, the resolution of the Hypothesis highlights the remarkable coherence of mathematics, where deep properties of primes, analytic functions, and asymptotic inequalities converge to settle one of the most profound questions in the discipline.


References

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  • [Nic22] Nicolas, J.-L. (2022). The sum of divisors function and the Riemann hypothesis. The Ramanujan Journal, 58, 1113–1157. https://doi.org/10.1007/s11139-021-00491-y

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MSC (2020): 11M26 (Nonreal zeros of ζ(s)\zeta (s) and L(s,χ)L(s, \chi) ); Riemann hypothesis), 11A25 (Arithmetic functions; related numbers; inversion formulas), 11A41 (Primes), 11N37 (Asymptotic results on arithmetic functions)


Documentation

Available as PDF at From Chebyshev to Primorials: Establishing the Riemann Hypothesis.

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