For a prime (or prime power) $p$ and some absolute constant $C$ (say $C$ = 100), consider the set $A$ of all $1 \leq a \leq p/C$ such that $1 \leq a^2 \leq p/C$ modulo $p$. Is it known that $|A| = \Omega(p)$?
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Yes. The points $(\frac{a}p,\frac{a^2\pmod p}p)$ are asymptotically equidistributed in $[0,1]^2$ by Weyl's criterion.
answered Aug 21, 2020 at 13:43
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$\begingroup$ Thanks a lot! Could you please give a reference to this result? $\endgroup$ Commented Aug 21, 2020 at 13:50
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1$\begingroup$ Read about Weyl's criterion here en.wikipedia.org/wiki/Equidistributed_sequence $\endgroup$ Commented Aug 21, 2020 at 14:41
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1$\begingroup$ And why is it applicable here: en.wikipedia.org/wiki/Quadratic_Gauss_sum $\endgroup$ Commented Aug 21, 2020 at 14:43
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2$\begingroup$ Of course one needs a multidimensional Weyl criterion here, see e.g. Prop. 1 at terrytao.wordpress.com/2010/03/28/… $\endgroup$ Commented Aug 21, 2020 at 19:58