Precise Control Over The Number Of Lattice Points

Let $D(P,Q)$ denote the distance between points $P$ and $Q$ in the Cartesian plane.

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Consider a point $P (\alpha , \beta )$ whose coordinates are distinct and both irrational.

Claim: The distance from $P$ to all lattice points are distinct.

Proof: Let $M (p, q)$ and $N( r,s)$ be two lattice points. We have to show that $D (P, M) = D(P,N) \implies M= N.$ Indeed, note that $D(P,M) = D(P,N) \implies (\alpha - p)^2 + (\beta - q)^2 = (\alpha - r)^2 + (\beta - s)^2 \\ \implies \alpha (2p - 2r) + \beta (2q - 2s) = p^2+q^2-r^2-s^2.$ Since $\alpha, \beta$ are distinct irrational numbers, for the LHS to be an integer, we must have $2p - 2r= 0, 2q- 2s= 0 \implies p=q, q=s \implies M=N,$ as desired. $\blacksquare$

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Now, let $T_n$ denote the $n^{\text{th}}$ closest point to $P.$ Consider any circle centered at $P$ and radius strictly between $D(P,T_{2014})$ and $D(P,T_{2015}).$ This circle contains precisely $2014$ lattice points in its interior. So, for all points with distinct irrational coordinates, we can find a circle with exactly $2014$ lattice points in its interior. There exist infinitely many points with distinct irrational coordinates which are pairwise rivals (for an explicit construction, consider $\{ (\sqrt{2} + i, \sqrt{3} + i) \}_{i=1}^{\infty}$ ). Therefore, our answer is $\boxed{\infty}.$