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Prime-power recursion S(ph,a)=pk−1S(ph−k,a)S(p^h,a)=p^{k-1}S(p^{h-k},a)S(ph,a)=pk−1S(ph−k,a)

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CMSharp.S_prime_pow_rec

by tabbott · Sep 2, 2026 · Mathlib c5ea003 (Lean v4.30.0)

analytic-number-theorycircle-methodexponential-sumsnumber-theory

Let ppp be a prime, k≥1k\ge1k≥1, and aaa an integer with p∤ap\nmid ap∤a. For the complete exponential sum

S(q,a)=∑m<qe ⁣(amkq),e(θ)=e2πiθ,S(q,a)=\sum_{m<q}e\!\left(\frac{am^{k}}{q}\right),\qquad e(\theta)=e^{2\pi i\theta},S(q,a)=m<q∑​e(qamk​),e(θ)=e2πiθ,

the value at a prime power reduces to a smaller one:

S(ph,a)  =  p k−1 S ⁣(p h−k,a)(k≤h).S(p^{h},a)\;=\;p^{\,k-1}\,S\!\left(p^{\,h-k},a\right)\qquad (k\le h).S(ph,a)=pk−1S(ph−k,a)(k≤h).

The hypothesis controlling the range is the existence of j≥1j\ge1j≥1 with pj∤kp^{j}\nmid kpj∤k and 2j≤h2j\le h2j≤h; for p∤kp\nmid kp∤k one may take j=1j=1j=1, so the recursion holds for all h≥max⁡(2,k)h\ge\max(2,k)h≥max(2,k).

The identity comes from splitting the sum by whether p∣mp\mid mp∣m. The part over mmm coprime to ppp vanishes, and the part over m=pm′m=pm'm=pm′ reindexes to pk−1S(ph−k,a)p^{k-1}S(p^{h-k},a)pk−1S(ph−k,a).

Iterating this descent is what produces the classical bound ∣S(q,a)∣≪q1−1/k+ε|S(q,a)|\ll q^{1-1/k+\varepsilon}∣S(q,a)∣≪q1−1/k+ε: the exponents match exactly, since pk−1⋅p(h−k)(1−1/k)=ph(1−1/k)p^{k-1}\cdot p^{(h-k)(1-1/k)}=p^{h(1-1/k)}pk−1⋅p(h−k)(1−1/k)=ph(1−1/k), so no saving is lost at any step.

Preamble
import Definitions.Def_CircleMethod_waring
import Mathlib.Data.Nat.Prime.Basic
Formal statement
namespace CMSharp

theorem S_prime_pow_rec {k p h j : ℕ} (hp : p.Prime) (a : ℤ) (ha : ¬ ((p : ℤ) ∣ a))
    (hj : 1 ≤ j) (hjh : 2 * j ≤ h) (hk : ¬ ((p : ℤ) ^ j ∣ (k : ℤ)))
    (hk1 : 1 ≤ k) (hkh : k ≤ h) :
    CircleMethod.S k (p ^ h) a = (p : ℂ) ^ (k - 1) * CircleMethod.S k (p ^ (h - k)) a := by sorry

end CMSharp
Source
R. C. Vaughan, The Hardy-Littlewood Method, 2nd ed., Cambridge Tracts in Mathematics 125, Cambridge University Press, 1997, Chapter 4, Theorem 4.2 and its proof in Section 4.1.

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