Skip to main content
7 events
when toggle format what by license comment
May 17, 2023 at 16:14 history edited Martin Sleziak
edited tags
May 17, 2023 at 16:10 answer added Maciej Ulas timeline score: 5
May 10, 2021 at 14:58 comment added Zhi-Wei Sun I conjecture further that for any prime $p>828$ there is a positive integer $a\le\sqrt{p-2}$ such that $a^2-1,2a,a^2+1$ are all quadratic residues modulo $p$. Note that $(a^2-1)^2+(2a)^2=(a^2+1)^2$.
May 9, 2021 at 15:22 comment added Zhi-Wei Sun By Fermat's last theorem, there are no positive squares $a,b,c$ satisfying $a^2+b^2=c^2$. This makes the question particularly interesting.
May 8, 2021 at 15:14 comment added Dr. Pi The circle method does not work in three variables, at least when you have congruences. See the Crelle paper (new version of circle method) of Heath-Brown (comments after Theorem 8).
May 7, 2021 at 11:06 comment added George Shakan For large primes you can use the circle method. See Corollary 4.15 in Tao and Vu's book on additive combinatorics for an easier variant, which may be modified.
May 6, 2021 at 22:02 history asked Zhi-Wei Sun CC BY-SA 4.0