# The clue to the proof of Riemann hypothesis

## Main Question or Discussion Point

This could be the way to proof. remember, this is not a proof.

today I found a clue to solution to Riemann hypothesis:

Let it be Riemann zeta function :ζ(s)

The proof that all the non trivial zeroes lie on the critical strip when s = 1/2 + it

let us suppose there are other zeroes having the real part between 0 < Re < 1/2

since to sattisfy the convergence of Re between 0 and 1/2 we know that THE VERY SAME AMOUNT OF ZEROS would accour symetrically to Re = 1/2

so the zeroes would accure on Re(1/2 - x) and they would occure on Re = (1/2 + x) because of convergence condition.

since the 2 values of Riemann Zeta functions of applying to it 2 different s1 and s2 s1 = 1/2 - X +b*i and s2 = 1/2 + X +b*i never equal, that means they couldnt have the same zeroes symetrically distributed to the Re = 1/2

---------------> the zeroes occur at Re=(1/2) if they exist.

The end.

Related Linear and Abstract Algebra News on Phys.org
If this is of any help to somebody, I would be glad, if not I appologize. I will take now some days off of Math problems. And yes, there is a great possibility, I am wrong... But I feel really passionate about Math and I think I will go to study theoretical Math. Greetings to all.

Revised :

In order to sattisfy DIVERGENCE criteria on interval 1> Re > 0 the zeroes are distributed simetrically to the Re = 1/2. Sorry I mixed the terms ...

REVISED:This could be the way to proof. remember, this is not a proof.

today I found a clue to solution to Riemann hypothesis:

Let it be Riemann zeta function :ζ(s)

The proof that all the non trivial zeroes lie on the critical strip when s = 1/2 + it

let us suppose there are other zeroes having the real part between 0 < Re < 1/2

since to sattisfy the divergence criteria of Re between 0 and 1 we know that THE VERY SAME AMOUNT OF ZEROS would accour on the right of Re=1/2 symetrically to Re = 1/2

so the zeroes would accure on Re(1/2 - x) and they would occure on Re = (1/2 + x) because of divergence condition.

since the 2 values of Riemann Zeta functions of applying to it 2 different s1 and s2 s1 = 1/2 - X +b*i and s2 = 1/2 + X +b*i never equal, that means they couldnt have the same zeroes symetrically distributed to the Re = 1/2

---------------> the zeroes occur at Re=(1/2) if they exist.