English
If χ is quadratic and χ(a) ≠ 1 for some a, then (χ(-1) · |R|)^{(p^n)/2} = χ(p^n) (in appropriate ring).
Русский
Если χ квадратичен и существует a с χ(a) ≠ 1, то (χ(-1) · |R|)^{(p^n)/2} = χ(p^n).
LaTeX
$$$\big( χ(-1) \cdot |R| \big)^{\frac{p^n}{2}} = χ(p^n)$$$
Lean4
theorem harmonic_eq_sum_Icc {n : ℕ} : harmonic n = ∑ i ∈ Finset.Icc 1 n, (↑i)⁻¹ :=
by
rw [harmonic, Finset.range_eq_Ico, Finset.sum_Ico_add' (fun (i : ℕ) ↦ (i : ℚ)⁻¹) 0 n (c := 1)]
simp only [Finset.Ico_add_one_right_eq_Icc]