Lattice points in translated discs

DefinitionLattice points in a translated unit disc#✓fully proved

For \(w = (w_1, w_2) \in \mathbb{R}^2\) and a natural number \(n\), the disc lattice count is the number of integer points \((p_1, p_2) \in \mathbb{Z}^2\) in the closed disc of radius \(1\) centred at \(n w\), that is, with \((p_1 - n w_1)^2 + (p_2 - n w_2)^2 \le 1\).

noncomputable def DiscLatticeCount (w : ℝ × ℝ) (n : ℕ) : ℕ :=
  Set.ncard {p : ℤ × ℤ | ((p.1 : ℝ) - (n : ℝ) * w.1) ^ 2 +
    ((p.2 : ℝ) - (n : ℝ) * w.2) ^ 2 ≤ 1}
Used by 1

TheoremSteinhaus's disc frequencies exist#●ready to prove

Let \(w = (w_1, w_2) \in \mathbb{R}^2\) be such that \(w_1\), \(w_2\) and \(w_1 / w_2\) are all irrational, and for each natural number \(n\) let \(c(n)\) be the disc lattice count of \(w\) and \(n\), the number of integer lattice points in the closed disc of radius \(1\) centred at \(n w\). Then for each \(m\) equal to \(2\), \(3\) or \(4\), the proportion of the integers \(n\) with \(0 \le n < N\) and \(c(n) = m\) converges to a real number as \(N \to \infty\).

A closed disc of radius \(1\) always covers at least two and at most five lattice points. The frequency values themselves are not pinned down here: they are the measures of the corresponding regions for the limiting distribution of \(n w\) modulo the lattice, and in particular depend on \(w\).

theorem_wanted steinhausDiscFrequenciesExist :
    ∀ w : ℝ × ℝ, Irrational w.1 → Irrational w.2 → Irrational (w.1 / w.2) →
      ∀ m : ℕ, m = 2 ∨ m = 3 ∨ m = 4 →
        ∃ freq : ℝ, Filter.Tendsto
          (fun N : ℕ => ((Finset.filter (fun n => DiscLatticeCount w n = m)
            (Finset.range N)).card : ℝ) / (N : ℝ))
          Filter.atTop (nhds freq)
Sources
  • H. Steinhaus, Problem 152 of the Scottish Book; J. Mycielski's commentary in Mauldin (ed.), The Scottish Book, 2nd ed. (2015) explains the solution using facts from Diophantine approximation, as reported in the 2026 survey "90+ years of the Scottish Book" (arXiv:2603.27867).
Uses 1
1 fully proved1 ready to prove

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