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Jan 8, 2010 at 21:40 comment added engelbrekt By the way, it is not necessary to know that $\zeta(s)$ has any zeros on the critical line, just that it has at least one zero in the critical strip $0 \leq \sigma \leq 1$. For then it has at least two zeros in $\sigma \geq 1/2$ by the functional equation and symmetry about the real axis, and only a single zero in $\sigma \geq 1/2$ is needed.
Jan 7, 2010 at 23:56 comment added gowers I was fairly sure that partial sums of mu were not better than the square root of n, but I didn't in fact know this argument, so thanks for giving it. I'll think about whether it can be adapted to work for the Liouville function.
Jan 7, 2010 at 23:23 history answered David E Speyer CC BY-SA 2.5