Skip to main content
8 events
when toggle format what by license comment
Nov 19 at 23:14 comment added GH from MO The first paragraph is not in harmony with the second paragraph. Specifically, the fact that the real number $\sqrt{x(x+1)}$ lies between $x$ and $x+1$ (for $x$ nonnegative) does not seem to help in solving quadratic equations of the form $x^2+x=b^2$ in real numbers. At any rate, assuming GRH (which is okay for pratical purposes), there is a fast algorithm for taking square-roots modulo $p$, hence also for solving quadratic equations modulo $p$.
Nov 19 at 19:00 comment added Turbo @sendit clarified.
Nov 19 at 19:00 history edited Turbo CC BY-SA 4.0
added 17 characters in body
Nov 19 at 18:53 comment added sendit The motivation for a ceil/floor function is unclear to me. In any case there isn't always a solution to x^2=b mod p. For example x^2=2 mod 3 has no solutions.
Nov 19 at 18:08 history edited Turbo CC BY-SA 4.0
added 194 characters in body
Nov 19 at 18:07 history undeleted Turbo
Nov 19 at 16:57 history deleted Turbo via Vote
Nov 19 at 16:55 history asked Turbo CC BY-SA 4.0