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In page 6, RH Equivalence 5.3. An equivalence of the Riemann Hypothesis says that

$$\sum_{\rho} \frac{1}{|\rho|^2} =\sum_{\rho} \frac{1}{\rho (1{-}\rho)}= 2 + \gamma - \log 4\pi$$ where $\rho$ is over nontrivial zeros of the Riemann zeta function. It's not hard to see that RH implies $$\sum_{\rho} \frac{1}{|\rho|^2}= 2 + \gamma - \log 4\pi.$$ But conversly I don't see how the equality above implies RH and there is no reference in page 6, RH Equivalence 5.3.

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Note that if $1/2< \sigma <1, t \in \mathbb R$ one has $\frac{2\sigma-1}{\sigma^2+t^2} < \frac{2\sigma-1}{(1-\sigma)^2+t^2}$.

By a little manipulation, one gets:

$\frac{2\sigma}{\sigma^2+t^2} + \frac{2(1-\sigma)}{(1-\sigma)^2+t^2} < \frac{1}{\sigma^2+t^2} + \frac{1}{(1-\sigma)^2+t^2} $

But if RH is false and there is $\rho=\sigma+it, 1/2<\sigma<1$, the above gives that

$2\Re{\frac{1}{\rho}}+2\Re{\frac{1}{1-\rho}}=2\Re{\frac{1}{\rho (1-\rho)}} < \frac{1}{|\rho|^2}+\frac{1}{|1-\rho|^2}$, so if we group together the four terms $\frac{1}{\rho (1-\rho)}, {\frac{1}{\bar \rho (1-\bar \rho)}}$ corresponding to the four roots $\rho, 1-\rho, \bar \rho, 1-\bar \rho$ we get that their sum is srictly less than the sum of the corresponding reciprocal of the respective four roots square modulus, so RH false implies $\sum_{\rho} \frac{1}{|\rho|^2} >\sum_{\rho} \frac{1}{\rho (1{-}\rho)}$ and the equivalence is established

Edit later - per comments - note that if $\Re \rho =1/2$ then $\bar \rho=1-\rho$ so roots group naturally in pairs only and $\frac{1}{\rho (1-\rho)}=\frac{1}{|\rho|^2}$ so the corresponding (two) terms on both sides of the equality $\sum_{\rho} \frac{1}{|\rho|^2} =\sum_{\rho} \frac{1}{\rho (1{-}\rho)}$ are equal

When there is a root with $\Re \rho_0 >1/2, t>0$ the roots group into four as noted $\rho_0, \bar \rho_0, 1-\rho_0, 1-\bar \rho_0$ and now the corresponding terms in $\sum_{\rho} \frac{1}{|\rho|^2}$ are $2(\frac{1}{|\rho_0|^2}+\frac{1}{|1-\rho_0|^2})$, while the terms in the sum $\sum_{\rho} \frac{1}{\rho (1{-}\rho)}$ are also four and since they are conjugate in pairs add to $2(2\Re{\frac{1}{\rho}}+2\Re{\frac{1}{1-\rho}})=4\Re{\frac{1}{\rho (1-\rho)}}$ and the inequality above applies

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    $\begingroup$ It's actually enough to group $\rho$ and $1-\rho$, because $\frac{1}{|\rho|^2}+\frac{1}{|1-\rho|^2}\geq 2\mathrm{Re}\,\frac{1}{\rho(1-\rho)}$ and equality is attained iff $\mathrm{Re}\,\rho=\frac12$ $\endgroup$
    – Alexander Kalmynin
    Commented Aug 1, 2021 at 15:17
  • $\begingroup$ It seems that the argument is not complete since it may happen that some of $\rho$ have real part 1/2 and some with $\sigma\neq 1/2$. In Conrad's answer it assumes that all $\rho$ have real part $\sigma\neq 1/2$. $\endgroup$
    – Beta
    Commented Aug 1, 2021 at 16:25
  • $\begingroup$ And also I'm a bit confused on how to group together. Could you please show the detailed calculation? Thanks very much! $\endgroup$
    – Beta
    Commented Aug 1, 2021 at 16:27
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    $\begingroup$ No, it is enough for one root to have real part greater than $1/2$ as for roots with real part $1/2$ trivially $1-\rho=\ bar \rho$ so the corresponding terms which now appear only in pairs are equal; $\endgroup$
    – Conrad
    Commented Aug 1, 2021 at 17:30

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