I expect we are far away...

How close to an answer sounds foreign to Math.

The right question should be is the Riemann Hypothesis proven or disproven? It is neither proven nor disproven so far.

michael atiyah thought, he had solved it. but he was wrong. the solution might fall literally from heaven.

Dear Alip Mohammed ,

1- The question “is the Riemann Hypothesis proven or disprove?” is a useless question because if it was proven, this fact would become extremely well know and all mathematicians would not that this Conjecture was proven/disproven and we would not even need to ask.

2- The question “how close are we to a solution for the Riemann Hypothesis?” is not foreign to mathematics as Mathematicians are often able to tell if we seem to be getting closer to a solution or not. For example, although a lot of work has been done on the Collatz conjecture, mathematicians can tell that we are not closer to a solution. Similarly, a few years ago, when Terrence Tao was asked about the Riemann Hypothesis, he said that unless a breakthrough happens he would not attempt it.

Dear Hinnerk Albert ,

I have heard of atiyah’s story. He has many great achievement. He beleived that he had solved it and that the system was wrongfully rejecting his proof. I which when I proof is found to be wrong, the reasons would be made public so that other researchers can easily know why a proof is wrong. Also this way authors cannot claim that they have been treated unfairly. The issue that journal often reject such proofs with any logical explanation, just stating that the proof is wrong. Well if the proof is wrong please at least tell the author what mistake you found, maybe he could then actually fixe the proof. Also most attempts at proof this conjecture are terrible and have some easy to stop mistakes. Everybody wants the famous that would come with solving this problem (and maybe the 1 000 000 $ cash prize also).

Dear Roudy El Haddad

A good blog explaining the interesting connections made in Riemann Hypothesis and the current breakthroughs is this

http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/riemannhyp.htm

Up until this point we have not seen any real and complete attempts to solve this hypothesis, but rather ideas that did not lead to a complete logical solution.

Dear Majid K.Neamah ,

I agree with you. Although year after year we are getting closer. A few year ago, it was proven that at least 40% of zeros have to be on the critical line. So a lot of progress has been done. Many conjecture have also been presented, which if proven, they would imply the Riemann Hypothesis. So have gotten some important result, however, we have not quite solved it yet.

Dear Ahmed Mohammed Ali ,

Interesting opinion. Can you share with us the reason that make you believe that we are still far away from a solution?

Lots of purported proofs (and some disproofs almost monthly) are frequently posted in arxiv general math and vixra, but their truth is still questioned.

A good elementary strategy would be to use Sondow's connection of highly composite numbers.

Hi Roudy!!!

I have made a tiny progress with 4 short notes in:

https://uh-cu.academia.edu/FrankVega

I will shortly introduce them:

1- Another Criterion For The Riemann Hypothesis.

Summary: Let's define $\delta(x) = (\sum_{{q\leq x}}{\frac{1}{q}}-\log \log x-B)$, where $B \approx 0.2614972128$ is the Meissel-Mertens constant. The Robin theorem states that $\delta(x)$ changes sign infinitely often. We define the another function $\varpi(x) = \left(\sum_{{q\leq x}}{\frac{1}{q}}-\log \log \theta(x)-B \right)$, where $\theta(x)$ is the Chebyshev function. We show that when the inequality $\varpi(x) \leq 0$ is satisfied for some number $x \geq 3$, then the Riemann Hypothesis should be false.

2- The Riemann Hypothesis.

Summary: For every prime number $p_{n}$, we define the sequence $X_{n} = \frac{\prod_{q \mid N_{n}} \frac{q}{q-1}}{e^{\gamma} \times \log\log N_{n}}$, where $N_{n} = \prod_{k = 1}^{n} p_{k}$ is the primorial number of order $n$ and $\gamma \approx 0.57721$ is the Euler-Mascheroni constant. We show if the sequence $X_{n}$ is strictly decreasing for $n$ big enough, then the Riemann hypothesis should be true. Moreover, we demonstrate that the sequence $X_{n}$ is indeed strictly decreasing when $n \to \infty$.

3- Robin Criterion on Divisibility

Summary: Robin criterion states that the Riemann Hypothesis is true if and only if the inequality $\sigma(n) < e^{\gamma } \times n \times \log \log n$ holds for all $n > 5040$, where $\sigma(n)$ is the sum-of-divisors function and $\gamma \approx 0.57721$ is the Euler-Mascheroni constant. This is known as the Robin inequality. In 2007, Choie, Lichiardopol, Moree and Sol{\'{e}} have shown that the Robin inequality is true for all $n > 5040$ which are not divisible by $2$. We prove that the Robin inequality is true for all $n > 5040$ which are not divisible by any prime number between $3$ and $953$.

4- The Robin Inequality On Certain Numbers

Summary: Robin criterion states that the Riemann Hypothesis is true if and only if the inequality $\sigma(n) < e^{\gamma } \times n \times \log \log n$ holds for all $n > 5040$, where $\sigma(n)$ is the sum-of-divisors function and $\gamma \approx 0.57721$ is the Euler-Mascheroni constant. This is called as the Robin inequality. Let $\free(n)$ denotes the square free kernel of a natural number $n$. Hence, we have that the Robin inequality holds when $ \frac{\pi^{2}}{6} \times \log\log \free(n) \leq \log\log n$ for some $n>5040$. Moreover, the Robin inequality holds for some $n>5040$ if every prime $q$ that divides $n$ satisfies $q < \sqrt[\beta]{\log n}$. Here, it is $\beta = e^{\gamma}$.

That's all,

I hope you could find them interesting,

Frank

Thank you dear Frank Vega . I will look into it when I find the time.

I put together a solution of the RH myself. While it can't be considered a complete proof while not vetted by experts, it presents various strong arguments and a real breakthrough, which is the inversion formula for Dirichlet series. Given any Dirichlet F(s), you know a(n) from F(s). Unfortunately, it's impossible to have an integral representation for a(n) usually, it's a Taylor power series. Please head to my page for the paper.