If RH is false , could this be true?

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 a funny quirk of mathematical logic that if the RH is true, then the answer to your question is yes. Thus nobody can answer no without disproving the RH first. So perhaps your question should be interpreted as “is it known that one cannot have more than two zeros in the critical strip with the same imaginary value?”. – Harald Hanche-Olsen Oct 8 '12 at 13:46
That is not true. If I would have asked " is it known that one cannot have more than 2 zeros in the critical strip with the same imaginary value " , the anwer would be " yes of course , we know this from RH + vote close or vote down or the comment that I should have added assuming RH is false IF THATS WHAT I MEANT. Now I did that from the start. And I do not ask for a proof of RH or a disproof. – mick Oct 8 '12 at 17:47
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. (Sorry, I am unable to parse the bulk of your comment, so I cannot respond.) – Harald Hanche-Olsen Oct 8 '12 at 20:03