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According to a theorem by H. Steinhaus, for every positive integer 'n' there is a circle in the 2 dimensional plane containing exactly 'n' lattice points in its interior. Can someone prove this theorem? As a teacher, I have been intrigued by this theorem.

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Here's the proof given by Ross Honsberger in Mathematical Gems:

The theorem follows easily from the fact that no two lattice points are the same distance from the point $(\sqrt2,1/3)$. We prove this shortly. Because of this, the lattice points can be ordered in a sequence $p_1,p_2,p_3,\dots$, according to their distance from $(\sqrt2,1/3)$; $p_1$ is closest, $p_2$ is next closest, and so on. Then the circle with center $(\sqrt2,1/3)$ which passes through $p_{n+1}$ contains in its interior precisely the $n$ lattice points $p_1,p_2,\dots,p_n$.

Suppose that the two lattice points $(a,b)$ and $(c,d)$ are the same distance from the point $(\sqrt2,1/3)$. Then $$(a-\sqrt2)^2+(b-(1/3))^2=(c-\sqrt2)^2+(d-(1/3))^2$$ Separating rational and irrational parts, we obtain $$2(c-a)\sqrt2=c^2+d^2-a^2-b^2+{2\over3}(b-d)$$ The left side is irrational or zero while the right side is rational. For equality both sides must be zero. Thus $$c=a,{\rm\ and\ }c^2+d^2-a^2-b^2+{2\over3}(b-d)=0$$ Since $c=a$, the latter gives $d^2-b^2-{2\over3}(d-b)=0$, and $$(d-b)(d+b-{2\over3})=0$$ Since $d$ and $b$ are integers, $d+b-(2/3)\ne0$. Thus we must have $d-b=0$, giving $d=b$. Since $c=a$, the two lattice popints $(a,b)$ and $(c,d)$ must be the same point, contradiction.

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The proof goes by induction where in the inductive step we shift and enlarge slightly the circle that worked in the previous step.

There are two extensions of this result to more general sets and more general spaces:

P. Zwoleński, Some generalization of Steinhaus’ lattice points problem, Colloq. Math. 123 (2011), 129–132.

T. Kania, T. Kochanek, Steinhaus’ lattice-point problem for Banach spaces, J. Math. Anal. Appl. 446 (2017), 1219–1229.

Note that the above-mentioned results give you dense sets of points such that cocentric circles with centres in a fixed point from such sets do the job.

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    $\begingroup$ Thanks for the links, but can you explain this in terms of 2-Dimensional plane only since the links cover Hilbert Space which is a very abstruse concept. $\endgroup$ – Harsh Gupta Mar 12 '17 at 10:45
  • $\begingroup$ @RahulSharma, the two dimensional space is also a Hilbert space. $\endgroup$ – Tomek Kania Mar 12 '17 at 11:03

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