Let $p$ be a prime number. According to Davenport (Multiplicative Number Theory, page 137) Schur proved (Indeed he proved much more, but let consider the simplest case) $$ \max_{t}\left|\sum_{n\leq t}\left(\frac{n}{p}\right)\right|>\frac{\sqrt{p}}{2\pi}. $$ What can we say about $t$ which we obtain the maximum? In other words, can we find $t\gg p^{\frac{1}{2}+\varepsilon}$ such that $$ \left|\sum_{n\leq t}\left(\frac{n}{p}\right)\right|>\frac{\sqrt{p}}{2\pi}. $$

  • $\begingroup$ Are you sure you want to write $\gg$ (large $t$) rather than $\ll$? $\endgroup$ – fedja Dec 14 '11 at 0:37
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    $\begingroup$ Burgess proved the character sum is $O(p^{3/16} t^{1/2} \log p)$, so if it's $\gg p^{1/2}$ then $t \gg p^{5/8 - \epsilon}$. $\endgroup$ – Noam D. Elkies Dec 14 '11 at 0:44
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    $\begingroup$ Fedja@: Yes. I am sure I want $\gg$. $\endgroup$ – M.B Dec 14 '11 at 2:23
  • $\begingroup$ Even if it were possible to have $t \ll p^{1/2 + \epsilon}$ then $p-t$ would work too and satisfy $t \gg p^{1/2 + \epsilon}$ by a wide margin... $\endgroup$ – Noam D. Elkies Dec 14 '11 at 3:44

The character sum you ask about is $\gg \sqrt{p}$ for at least one value of $n$. I learned the following slick proof from a paper of Leo Goldmakher: We have

$$\tau(p) = \sum_{n \leq p} \bigg( \frac{n}{p} \bigg) e^{2 \pi i n/p}$$ and that is a Gauss sum with absolute value $\sqrt{p}$... now use partial summation.

In addition, there is an infinite family of characters for which the lower bound may be multiplied by an additional fractional power of $\log \log p$. A result like this was first proved by Paley. Please see the paper I linked to for proofs and references to earlier work.


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