From Chebyshev to Primorials: Establishing the Riemann Hypothesis

1. Introduction

The Riemann Hypothesis, first proposed by Bernhard Riemann in 1859, asserts that all non-trivial zeros of the Riemann zeta function $\zeta \left(s\right)$ lie on the critical line $\Re \left(s\right)={\displaystyle \frac{1}{2}}$. 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 [1].

The zeta function $\zeta \left(s\right)$, 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 ${\displaystyle \frac{1}{2}}$. 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.

1.1. Main Result

In this work, we establish the hypothesis by introducing a criterion based on the comparative growth of Chebyshev’s $\theta$-function and primorial numbers. Specifically, we show that for every sufficiently large prime $p_n$, there exists a larger prime $p_{n^{\prime }}$ such that the ratio $R\left(N_{n^{\prime }}\right)$, defined via the Dedekind $\mathsf{\Psi }$-function and primorials, satisfies $R\left(N_{n^{\prime }}\right)<R\left(N_n\right)$.

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

$${\displaystyle \frac{log\left(\theta \left(p_{n^{\prime }}\right)\right)}{log\left(\theta \left(p_n\right)\right)}}>\prod _{p_n<p\le p_{n^{\prime }}}\left(1+{\displaystyle \frac{1}{p}}\right).$$

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 $\mathsf{\Psi }$ function play a central role.

2.1. The Chebyshev Function

The Chebyshev function $\theta \left(x\right)$ is defined by

$$\theta \left(x\right)=\sum _{p\le x}logp$$

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

2.2. The Riemann Zeta Function

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

$$\zeta \left(2\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{n^2}}.$$

Proposition 1. 

The value of the Riemann zeta function at $s=2$ satisfies

$$\zeta \left(2\right)=\prod _{k=1}^{\infty }{\displaystyle \frac{p_k^2}{p_k^2-1}}={\displaystyle \frac{{\pi }^2}{6}}$$

where $p_k$ denotes the k-th prime number [2].

2.3. The Dedekind $\mathsf{\Psi }$ Function and Primorials

For a natural number n, the Dedekind $\mathsf{\Psi }$ function is defined as

$$\mathsf{\Psi }\left(n\right)=n·\prod _{p\mid n}\left(1+{\displaystyle \frac{1}{p}}\right)$$

where the product runs over all prime divisors of n.

The k-th primorial, denoted $N_k$, is

$$N_k=\prod _{i=1}^k{p}_i$$

the product of the first k primes.

We further define, for $n\ge 3$:

$$R\left(n\right)={\displaystyle \frac{\mathsf{\Psi }\left(n\right)}{n·loglogn}}.$$

For the n-th prime $p_n$, we say that the condition $\mathsf{Dedekind}\left(p_n\right)$ holds if

$$\prod _{p\le p_n}\left(1+{\displaystyle \frac{1}{p}}\right)>{\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}·log\theta \left(p_n\right)$$

where $\gamma$ is the Euler—Mascheroni constant. Equivalently, $\mathsf{Dedekind}\left(p_n\right)$ holds if and only if

$$R\left(N_n\right)>{\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}.$$

Proposition 2. 

If the Riemann Hypothesis is false, then there exist infinitely many n such that [3]:

$$R\left(N_n\right)<{\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}.$$

Proposition 3. 

As $k\to \infty$, the sequence $R\left(N_k\right)$ converges to [4]:

$$\underset{k\to \infty }{lim}R\left(N_k\right)={\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}.$$

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 $\alpha >1$ be fixed. Then there exists $N\in \mathbb{N}$ such that for all $n>N$ there is an integer i with

$${\displaystyle \frac{log\theta \left(p_{n+i}\right)}{log\theta \left(p_n\right)}}>\prod _{p_n<p\le p_{n+i}}\left(1+{\displaystyle \frac{1}{p}}\right).$$

Proof. 

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

Step 1. Reduction of the product

We use the identity

$$\prod _{p_n<p\le p_{n+i}}\left(1+{\displaystyle \frac{1}{p}}\right)={\displaystyle \frac{{\prod }_{p_n<p\le p_{n+i}}\left(1-\frac{1}{p^2}\right)}{{\prod }_{p_n<p\le p_{n+i}}\left(1-\frac{1}{p}\right)}}.$$

Thus it suffices to prove

$${\displaystyle \frac{log\theta \left(p_{n+i}\right)}{log\theta \left(p_n\right)}}·\prod _{p_n<p\le p_{n+i}}\left(1-{\displaystyle \frac{1}{p}}\right)>\prod _{p_n<p\le p_{n+i}}\left(1-{\displaystyle \frac{1}{p^2}}\right).$$

Step 2. Choice of i

Fix $\alpha >1$. For each n, let i be chosen so that $p_{n+i}$ is the largest prime with

$$p_{n+i}\le p_n^{\alpha }.$$

As $n\to \infty$, this ensures $p_{n+i}\sim p_n^{\alpha }$.

Step 3. Growth of the logarithmic ratio

By the Prime Number Theorem, $\theta \left(x\right)\sim x$ [5]. Hence

$$\underset{n\to \infty }{lim}{\displaystyle \frac{log\theta \left(p_{n+i}\right)}{log\theta \left(p_n\right)}}=\underset{n\to \infty }{lim}{\displaystyle \frac{log{p}_{n+i}}{log{p}_n}}=\underset{n\to \infty }{lim}{\displaystyle \frac{log\left(p_n^{\alpha }\right)}{log{p}_n}}=\alpha .$$

Thus, for large n, this ratio is arbitrarily close to $\alpha$.

Step 4. Behavior of the Euler product factor

We rewrite

$$\prod _{p_n<p\le p_{n+i}}\left(1-{\displaystyle \frac{1}{p}}\right)={\displaystyle \frac{{\prod }_{p\le p_{n+i}}\left(1-\frac{1}{p}\right)}{{\prod }_{p\le p_n}\left(1-\frac{1}{p}\right)}}.$$

By Mertens’ theorem [6],

$$\prod _{p\le x}\left(1-{\displaystyle \frac{1}{p}}\right)\sim {\displaystyle \frac{e^{-\gamma }}{logx}}.$$

Therefore,

$$\underset{n\to \infty }{lim}{\displaystyle \frac{{\prod }_{p\le p_{n+i}}\left(1-\frac{1}{p}\right)}{{\prod }_{p\le p_n}\left(1-\frac{1}{p}\right)}}=\underset{n\to \infty }{lim}{\displaystyle \frac{log{p}_n}{log{p}_{n+i}}}={\displaystyle \frac{1}{\alpha }}.$$

So for large n, this product is arbitrarily close to $1/\alpha$.

Step 5. Contribution of the squared terms

From explicit bounds (see [7]), for $p_n>24317$ one has

$$\begin{array}{cc}\hfill & -{\displaystyle \frac{1}{p_n·log{p}_n}}+{\displaystyle \frac{1}{p_n·{log}^2{p}_n}}-{\displaystyle \frac{2}{p_n·{log}^3{p}_n}}+{\displaystyle \frac{2}{p_n·{log}^4{p}_n}}\hfill \\ \hfill & \le \sum _{p_n\le p}log\left(1-{\displaystyle \frac{1}{p^2}}\right)\hfill \\ \hfill & \le -{\displaystyle \frac{1}{p_n·log{p}_n}}+{\displaystyle \frac{1}{p_n·{log}^2{p}_n}}-{\displaystyle \frac{2}{p_n·{log}^3{p}_n}}+{\displaystyle \frac{10.26}{p_n·{log}^4{p}_n}}\hfill \end{array}$$

In particular,

$$\sum _{p_n<p\le p_{n+i}}log\left(1-{\displaystyle \frac{1}{p^2}}\right)\sim -{\displaystyle \frac{1}{p_{n}log{p}_n}}$$

as $i\to \infty$.

Step 6. Final comparison

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

$$log\left(\alpha ·{\displaystyle \frac{1}{\alpha }}\right)=0$$

while the right-hand side is asymptotic to $-1/(p_{n}log{p}_n)$, which is strictly negative. Hence, for sufficiently large n, the inequality holds.

Step 7. Conclusion

Thus, for every $\alpha >1$ there exists N such that for all $n>N$ the inequality is satisfied for the chosen i.

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Lemma 2 (Main Insight).

The Riemann Hypothesis holds provided that, for some sufficiently large prime $p_n$, there exists a larger prime $p_{n^{\prime }}>p_n$ such that

$$R\left(N_{n^{\prime }}\right)<R\left(N_n\right).$$

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 $R\left(N_k\right)$.

Step 1. Existence of a starting point

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

$$R\left(N_n\right)<{\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}.$$

Choose one such index $n_1$ corresponding to a prime $p_{n_1}$.

Step 2. Iterative construction

By the hypothesis of the lemma, whenever $R\left(N_n\right)<{\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}$ there exists a larger prime $p_{n^{\prime }}>p_n$ with

$$R\left(N_{n^{\prime }}\right)<R\left(N_n\right).$$

Applying this iteratively starting from $n_1$, we obtain an infinite increasing sequence of indices

$$n_1<n_2<n_3<\dots$$

such that

$$R\left(N_{n_{i+1}}\right)<R\left(N_{n_i}\right)\mathrm{for}\mathrm{all}i\ge 1.$$

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

Step 3. Contradiction with the limit

By Proposition 3, we know that

$$\underset{k\to \infty }{lim}R\left(N_k\right)={\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}.$$

Hence, for any $\epsilon >0$, there exists K such that for all $k>K$,

$$\left|R\left(N_k\right)-{\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}\right|<\epsilon .$$

Take

$$\epsilon ={\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}-R\left(N_{n_1}\right)>0.$$

By convergence, only finitely many terms of $\left\{R\left(N_k\right)\right\}$ can lie below ${\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}-\epsilon$. However, the subsequence $\left\{R\left(N_{n_i}\right)\right\}$ is infinite and satisfies

$$R\left(N_{n_i}\right)<{\displaystyle \frac{e^{\gamma }}{\zeta \left(2\right)}}-\epsilon \mathrm{for}\mathrm{all}i\ge 1$$

a contradiction.

Step 4. Conclusion

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

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Theorem 1 (Main Theorem).

The Riemann Hypothesis is true.

Proof.

By Lemma 2, the Riemann Hypothesis holds if, for some sufficiently large prime $p_n$, there exists a larger prime $p_{n^{\prime }}>p_n$ such that

$$R\left(N_{n^{\prime }}\right)<R\left(N_n\right).$$

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

Step 1. Expression for $R\left(N_k\right)$

For the k-th primorial $N_k={\prod }_{i=1}^k{p}_i$, we have

$$R\left(N_k\right)={\displaystyle \frac{\mathsf{\Psi }\left(N_k\right)}{N_{k}loglog{N}_k}}={\displaystyle \frac{{\prod }_{i=1}^k\left(1+\frac{1}{p_i}\right)}{loglog{N}_k}}.$$

Since $\theta \left(p_k\right)={\sum }_{i=1}^{k}log{p}_i=log{N}_k$, it follows that

$$loglog{N}_k=log\theta \left(p_k\right).$$

Thus,

$$R\left(N_k\right)={\displaystyle \frac{{\prod }_{i=1}^k\left(1+\frac{1}{p_i}\right)}{log\theta \left(p_k\right)}}.$$

Step 2. Reformulating the inequality

The condition $R\left(N_{n^{\prime }}\right)<R\left(N_n\right)$ is equivalent to

$${\displaystyle \frac{{\prod }_{i=1}^{n^{\prime }}\left(1+\frac{1}{p_i}\right)}{log\theta \left(p_{n^{\prime }}\right)}}<{\displaystyle \frac{{\prod }_{i=1}^n\left(1+\frac{1}{p_i}\right)}{log\theta \left(p_n\right)}}.$$

Rearranging gives

$${\displaystyle \frac{log\theta \left(p_{n^{\prime }}\right)}{log\theta \left(p_n\right)}}>{\displaystyle \frac{{\prod }_{i=1}^{n^{\prime }}\left(1+\frac{1}{p_i}\right)}{{\prod }_{i=1}^n\left(1+\frac{1}{p_i}\right)}}=\prod _{p_n<p\le p_{n^{\prime }}}\left(1+{\displaystyle \frac{1}{p}}\right).$$

Hence the inequality is equivalent to

$${\displaystyle \frac{log\theta \left(p_{n^{\prime }}\right)}{log\theta \left(p_n\right)}}>\prod _{p_n<p\le p_{n^{\prime }}}\left(1+{\displaystyle \frac{1}{p}}\right).$$

Step 3. Conclusion

By Lemma 1, this inequality holds for sufficiently large $p_n$. Therefore, for such $p_n$ there exists $p_{n^{\prime }}>p_n$ with $R\left(N_{n^{\prime }}\right)<R\left(N_n\right)$. By Lemma 2, this implies the Riemann Hypothesis.

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