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## 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.

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.