TITLE: Reference request: Proof of H. Steinhaus' Lattice-Point Theorem
QUESTION [5 upvotes]: 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.

REPLY [6 votes]: 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.