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Today I did some reading on the Riemann Hypothesis and decided to play around with $\zeta(s)$ a little bit. (In case my question is ridiculous, I'm a student who has no experience dealing with zeta functions - I've only ever dealt with their components.)

I was wondering if it would be possible that a lot of luck/creativity would allow one to simplify $\zeta(s)=0$, or an analytic continuation of $\zeta(s)$, and show that $\Re(s)=1/2$? Or would a proof have to involve some more abstract, qualitative ideas? If the latter, is it likely that there are (undiscovered) non-trivial ideas that would just require a really creative simplification?

I mean, I'm playing around with it because it's fun/good practice...but I'm still curious if there's a remote possibility that an amateur (i.e., someone with a math degree and some graduate courses under their belt) could find something.

Thanks for any feedback.

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The Prime Number Theorem was long thought by many number theorists (such as Hardy) to have no elementary proof. It was thought that the PNT was a "hard" and "deep" theorem and as it can be translated to say something about the Riemann zeta function, it should be unlikely one could prove the PNT without complex function theory.

However, a few decades after Hardy shared his thoughts, Erdos and Selberg produced elementary proofs that a sufficiently creative person with no knowledge of complex analysis could produce, so perhaps the answer to your question is that no one knows that it is impossible for a sufficiently creative person to find an elementary proof without very powerful tools, and because all the experts have tried and failed so far we are inclined to think, as Hardy did, that it is probably not possible. Perhaps time will come to show that a proof using only basic complex function theory exists.

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Erdos-Selberg is elementary in the technical sense of not using complex numbers, but not at all in the colloquial sense of being easy. It is, in my opinion, harder than the complex analysis proofs. Also my understanding is that the Erdos-Selberg proof is modeled on the complex variable proofs and probably couldn't have been discovered without those earlier proofs as a guideline. – Gerry Myerson Sep 8 '12 at 3:01
@Gerry I meant "elementary" in the same sense as you, I should have mentioned that in my post. I agree it is definitely harder than any of the classical proofs using complex analysis. I just wanted to convey that in the totally theoretical sense, I think it is indeed "possible that a lot of luck/creativity" could lead someone to discover Erdos/Selbergs proof (though of course extremely unlikely), so maybe a similar statement applies to RH as well. – Ragib Zaman Sep 8 '12 at 3:14
Thanks for the example. Now I've got to hunt that proof down - sounds like an interesting proof to go through. – James000 Sep 8 '12 at 3:19
Didn't realize you can't hit "enter" in a comment. Anyway, thanks for taking my question extremely theoretically. I really was just wanting to know if there was a $\epsilon$-chance that a simple method could lead to something interesting. But I should mention that I know an amateur would probably only find something if it fell from the heavens. And since we killed god, according to Nietzsche at least, I really doubt that will happen. :) – James000 Sep 8 '12 at 3:31

"I'm still curious if there's a remote possibility that an amateur (i.e., someone with a math degree and some graduate courses under their belt) could find something."

We have no way to know, until we get there and see what it actually takes. But some truly brilliant people have looked at RH, and have tried using a great many different approaches, and I'd have to say I'd bet against an amateur getting anything useful.

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there is a WAY using Quantum mechanics and the Abel's integral equation to get a partial result as i have pointed inside

in other words, use the Riemann-Weil trace formula , see :

and the potential (the inverse) of a Hamiltonian $$ -y''(x)+f(x)y(x) $$whose eigenvalues are the square of the Riemann zeros is given implicitly by

$$ f^{-1}= 2\sum_{n=0}^{\infty} \frac{H(x-\gamma_{n}^{2})}{\sqrt{\pi}(x-\gamma_{n}^{2})} $$

insert this into the Riemann weil formula and you will get the impicit expression for thepotential

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