View Single Post
Old 2021-11-16, 15:05   #3
RomanM
 
Jun 2021

3×17 Posts
Default

Thank You for the answer! The above arose from this (2m^{3}\pm1)^{2}\equiv1\; mod\;m^{5}\pm m^{2}
but m^5+-m^2 is narrow number to match any chosen p. If m=x/y; all much more interesting; if we took numerators of fractions that turned out by this substitution, the one (==1) turns to be square.

Last fiddled with by RomanM on 2021-11-16 at 15:06
RomanM is offline   Reply With Quote