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 2021-11-16, 15:05 #3 RomanM   Jun 2021 3×17 Posts 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