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Aug 28, 2018 at 10:14 comment added Watson K. Conrad also wrote an answer here related to these results.
Jun 5, 2016 at 20:23 vote accept CommunityBot
May 24, 2016 at 16:35 comment added David E Speyer See also Kuo and Murty mast.queensu.ca/~murty/Kuo-Murty-CJM.pdf
May 24, 2016 at 16:32 comment added David E Speyer In general, Keith's paper is very clear and pretty; I recommend it.
May 24, 2016 at 16:32 comment added David E Speyer Regarding the converse: Let $\alpha_p$ and $\beta_p$ be the Frobenius eigenvalues and set $\psi_E(x) = \sum_{p^k < x} (\alpha_p^k + \beta_p^k) \log p$. Keith Conrad math.uconn.edu/~kconrad/articles/eulerprod.pdf (Theorem 1.3) shows that $f_E(x) \sim C (\log x)^r$ is equivalent to (ii) together with $\psi_E(x) = o(x \log x)$, while GRH is equivalent to $\psi_E(x) = O(x (\log x)^2)$, so the Euler product condition appears stronger. (Of course, it is possible someone could find some bootstrapping argument which shows the $o(x \log x)$ and $O(x (\log x)^2)$ bounds are equivalent.)
May 24, 2016 at 16:27 history answered Jeremy Rouse CC BY-SA 3.0