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Let $\zeta(s)$ be the Riemann zeta function.

Assume RH is false , is it possible that we have in the critical strip $\zeta(a_1+ti) = \zeta(a_2+ti) = \zeta(a_3+ti) = \cdots = \zeta(a_n+ti) = 0$

For $a_n$ real and $t$ real and $n$ an integer > 2 ?

Can we put restrictions on the divisors of $n$ ?

In particular I wonder if $n=3$ is possible. I wonder because then we have by symmetry of the zero's also a real part $1/2$ ; $1/2+ti$ must then be a zero.

Does the multiplicity of the $\zeta$ zeroes have any effect on this ?

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    It *is* true of the question as stated in the title, but the body text is better. My comment was a bit tongue in cheek; I apologize if that wasn't clear. I never really believed you were asking about the RH itself, of course. ($S$orry, I am unable to parse the bulk of your comment, so I cannot respond.)2012-10-08

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