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I was wondering if the following could be established by the methods that go into e.g. Linnik:

$\textbf{Claim. } \text{Let $\chi$ be a nonprincipal quadratic character of conductor $q$, and (e.g.) $c := \frac{1}{100}$. Then $\sum_{p\leq q^c} (1+\chi(p))\geq q^\delta$ once $q\gg_\delta 1$.}$

Of course once $c\gg 1$ this follows from Linnik's theorem. I'm guessing the answer here is no, since the interval is so short, but I'm not sure since the required lower bound is fairly small. (Forgive me if I have overlooked a very easy obstruction!)

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    $\begingroup$ This would imply bounds on the least prime quadratic residue mod q that go beyond the current world record coming from the Burgess bound, so I'd say that this is beyond the reach of current technology (unless one assumes some very good lower bound on $L(1,\chi)$, such as that coming from GRH). See also mathoverflow.net/questions/137714/… $\endgroup$
    – Terry Tao
    Jul 9, 2015 at 21:30
  • $\begingroup$ Perfect! Thanks! It seems the lower bound needed to overcome that obstruction is just $L(1,\chi) > \frac{C}{\log{q}}$ for $C\gg 1$, by the result of Wolke mentioned in that link (or matwbn.icm.edu.pl/ksiazki/aa/aa32/aa3227.pdf). I'm happy to assume the character doesn't have a Siegel zero, in which case one gets the needed lower bound only for some (very small!) $C$. I wonder if just assuming the lower bound needed for Wolke is enough to get this quantitative form? ...Probably not, again. $\endgroup$
    – alpoge
    Jul 9, 2015 at 22:05

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