Timeline for On the parity of $|\{(j,k):\ 1\le j<k\le\frac{p-1}2\ \&\ \ j(j+1)\ \text{mod}\ p\,>\,k(k+1)\ \text{mod}\ p\}|$ with $p$ prime

11 events

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Oct 29, 2018 at 3:04	comment	added	Fedor Petrov		@Zhi-WeiSun Thank you, I added this reference to the answer but keep the proof for completeness.
Oct 29, 2018 at 2:33	history	edited	Fedor Petrov	CC BY-SA 4.0	added 153 characters in body
Oct 29, 2018 at 2:24	comment	added	Zhi-Wei Sun		For prime $p\equiv7\pmod 8$, Berndt and Chowla showed that $\sum_{k=\lfloor p/4\rfloor}^{\lfloor p/2\rfloor}(\frac kp)=0$. Note also that $\sum_{k=1}^{\lfloor p/2\rfloor}(\frac kp)=h(-p)$. So the case $p=8k-1$ has been solved as well.
Oct 29, 2018 at 2:21	comment	added	Zhi-Wei Sun		In 1974 B.C.Berndt and S. Chowla[Nordisk Mat. Tidskr. 22(1974), 5-8] proved that $\sum_{k=1}^{\lfloor p/4\rfloor}(\frac kp)=0$ for any prime $p\equiv3\pmod 8$. So the case $p=8k+3$ has been solved.
Oct 29, 2018 at 1:30	history	edited	Fedor Petrov	CC BY-SA 4.0	added 3 characters in body
Oct 29, 2018 at 1:27	comment	added	Zhi-Wei Sun		Okay, it's my negligence.
Oct 29, 2018 at 1:19	comment	added	Fedor Petrov		I think, $\{s^2 - A\} _p=p+\{s^2\} _p- A$ for $\{s^2\}_p<A$
Oct 28, 2018 at 23:07	comment	added	Zhi-Wei Sun		How do you get the equality for signs? If $\{s^2\}_p<A<\{t^2\}_p$ then $$\{s^2-A\}_p-\{t^2-A\}_p=A-\{s^2\}_p-(\{t^2\}_p-A)=\frac{p+1}2-\{s^2\}_p-\{t^2\}_p.$$
Oct 28, 2018 at 17:38	history	edited	Fedor Petrov	CC BY-SA 4.0	added 1006 characters in body
Oct 27, 2018 at 16:50	history	edited	Fedor Petrov	CC BY-SA 4.0	added 367 characters in body
Oct 27, 2018 at 15:14	history	answered	Fedor Petrov	CC BY-SA 4.0