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The question is the title. For example, if we could show that $S$ is finite, then this would entail that every large enough integer $n$ is such that $\zeta(2n+1)$ is irrational and that, under RH, almost all non-trivial zeros of $\zeta$ have irrational imaginary part. Has this question been studied? Any reference?
Thank you in advance.

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This includes all trivial zeros so it is infinite. –  joro Oct 12 '13 at 14:36
    
Is there another example except nonpositive integers? –  joro Oct 12 '13 at 14:37
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2 Answers 2

up vote 3 down vote accepted

This set includes the non-positive integers (including the trivial zeros) so the set is infinite.

I would be interested in another example.

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maybe you should not accept this answer because it is rather trivial? –  Sam Hopkins Dec 17 '13 at 16:14
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See D. Maser's Rational Values of the Riemann zeta function. In it he proves that the number of rational numbers $s\in[2,3]$ with denominator at most $D$ such that $\zeta(s)$ is also a rational with denominator at most $D$ is $O\left( \left(\frac{\log(D)}{\log\log(D)}\right)^2 \right)$.

Of course the set is probably empty.

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Is there a nontrivial solution known? –  joro Dec 18 '13 at 12:29
    
@joro, not as far as I know. –  Mark Lewko Dec 18 '13 at 18:27
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