Iterated Dickson polynomials over finite fields: when the eventual period is k and when it is k/2

Section 4 of the source – Lemma 4.4 (script-l = l), Proposition 4.5 (k odd implies script-k = k) and the k-even dichotomy script-k in k, k/2 – verified KERNEL-CLEAN (decide, axioms [propext]) over F₃, F₄ = F₂[X]/(X²+X+1) and F₅, for every alpha in F_qˣ and every n <= pi_q(alpha) with alphaⁿ = alpha. The checked Bool also certifies that the presented ring is a field (no zero divisors, hence p prime and the defining polynomial irreducible).

The same sweep by native_decide at fourteen more fields – q = 7, 8, 9, 11, 13, 16, 17, 19, 23, 25, 27, 29, 31, 32, covering both parities of the characteristic, prime and non-prime q, and square and non-square alpha – for 13,866 admissible (alpha, n) pairs in all (with DE1).

The source's five printed number blocks – Tables 1a (D₁20(x,-1) over F₁1, coefficients of x²... x¹20), 1b (3*D₁20(x,2)), 2 (Table 1a re-laid in eleven columns), 3a (D₆0(x,1), x⁶0 down to x²) and 3b (2*D₆1(x,9), x⁵9 down to x) – recomputed from the recurrence D₀ = 2, D₁ = x, Dₘ = x Dₘ₋₁ - alpha Dₘ₋₂ alone and matched cell for cell, kernel-clean; the 180-degree rotation of Table 3 checked as the equation table3a = reverse(table3b).

Theorem 1.1 of the source – the exact period of [Dₙ(x,alpha) mod (x^q - x)]_(n>=1) is (q²-1)/2 when q is odd and alpha is a square in F_q, and q²-1 otherwise – recomputed for EVERY alpha in F_qˣ at q = 3, 4, 5, 7, 8, 9, 11, 13, 16, from the exact-period definition (least t >= 1 with (D₁₊ₜ, D₂₊ₜ) = (D₁, D₂), which suffices because the two-term recurrence is invertible for alpha!= 0).

THE ANSWER TO THE SOURCE'S QUESTION. Let pi = pi_q(alpha), let pi₂ be the largest divisor of pi coprime to n (so k is the multiplicative order of n modulo pi₂), and inside (Z/pi₂ Z)ˣ let Kₐlpha = 1, q when alpha!= 1 and Kₐlpha = 1, -1, q, -q when alpha = 1. Then, for every alpha in F_qˣ and every n with alphaⁿ = alpha, script-k = k / |<n> cap Kₐlpha|. Equivalently, for k even: script-k = k/2 exactly when the unique involution n^(k/2) of the cyclic group <n> lies in Kₐlpha, and script-k = k otherwise. The NEW content is the non-permutation regime gcd(n, q²-1) > 1 at alpha!= 1 – precisely the regime the Question is posed in – together with the single global formula that unifies all four regimes. Kernel-bound at 17 fields q <= 32 (kernel-CLEAN at q = 3, 4, 5), 13,866 pairs.

How the Question's two branches split. Over alphaⁿ = alpha, gcd(n, q²-1) > 1 and k even, the pair (#script-k = k/2, #script-k = k) is (0,0) at q = 3, (4,0) at 4, (5,1) at 5, (6,8) at 7, (12,0) at 8, (17,7) at 9, (50,32) at 11, (66,32) at 13, (204,16) at 16, (51,69) at 17 and (186,108) at 19. Neither column is monotone in q (compare q = 31, where halving is rare, with q = 32, where it is the rule).

D_q(x, alpha) = x (mod x^q - x) for every alpha in F_qˣ, at q = 3, 4, 5, 7, 8, 9 – the identity that puts the Frobenius class q into Kₐlpha, since it gives D_(qm)(.,alpha) = Dₘ(.,alpha) whenever alphaᵐ = alpha.

Negative controls for DE5, each measured rather than asserted: (i) dropping the source's standing hypothesis alphaⁿ = alpha makes the k-even dichotomy itself FALSE – 2 pairs at q = 5 and 9 at q = 7 have script-k neither k nor k/2 (0 under the hypothesis at every field swept); (ii) dropping the classes -1 and -q from K₁ makes the criterion wrong 3 times at q = 5 and 6 times at q = 7; (iii) replacing pi_q(alpha) by q² - 1 makes it wrong 2 times at q = 5 and 10 at q = 7; (iv) Fld.isField refuses F₂[X]/(X²+1), so no sweep can run over a ring that is not a field.

Non-vacuity of the Question at the smallest field where it has both answers: over F₅ the split is (5, 1), refuting at once the too-small reading ('script-k = k always') and the too-large one ('script-k = k/2 whenever k is even'). F₃ has no admissible case with k even at all and F₄ has only halving ones, so F₅ is minimal.

Corollary 4.3(1) of arXiv:2508.08621v3 is printed with its delta_(n,alpha) condition INVERTED: as printed it sets delta = 1/2 when the kernel class is NOT a power of n, whereas the period halves exactly when it IS. Two kernel-clean witnesses over F₅, both with alpha = 2 (so pi₅(2) = 24): n = 5 has k = 2, true period script-k = 1, and q = 5 = n¹ mod 24 – the printed condition predicts 2; n = 13 has k = 2, true period script-k = 2, and q = 5 is not in <13> = 13, 1 mod 24 – the printed condition predicts 1. Both directions fail, which locates the defect in the condition rather than in an exponent.

The sufficiency half of DE5, PROVED (in prose, not formalized): if alphaⁿ = alpha, k is even and n^(k/2) = eps mod pi₂ for some eps in Kₐlpha other than 1, then script-k divides k/2. Proof: pi₁ divides nˡ, so n^(l+k/2) = eps * nˡ mod pi by CRT; then D_(n^(l+k/2)) = D_(eps nˡ) = D_(nˡ) using D_(qm)(.,alpha) = Dₘ(.,alpha) (row DE7) and, at alpha = 1, D₋ₘ(.,1) = Dₘ(.,1), together with the source's Theorem 1.1.

The attribution boundary of row DE5, made checkable. At alpha = 1 (where Dₙ(x,1) = 2 Tₙ(x/2) for odd q, so the functional graph is Chebyshev's up to conjugation) this family's script-k equals lcm over d | omega₀ or d | omega₁ of otilde_d(n), the aggregate of Qureshi-Panario's per-point period, at every odd prime q <= 19 and every n <= pi_q(1) – kernel-clean by decide at q = 3, 5, 7 and by native_decide at q = 11, 13, 17, 19. And script-k is nevertheless NOT a function of (q, n) alone: at q = 13, n = 3 the two squares alpha = 1 and alpha = -1 share pi_q(alpha) = 84 and give script-k = 3 and 6.

The Ω-restricted analogue of Chen–Peng's Lemma 4.2 — the lemma the founding's §8 named as the gap — stated and proved in complete generality and kernel-clean: for a commutative group G, finite subgroups P, R ≤ G, w ∈ G, integers Q, s with u^(Q−1) = 1 on P, h^(Q+1) = 1 on R, s coprime to |P| and |R|, and E generating the annihilator of P ∪ wR: every point of P ∪ wR satisfies u^(s−1) = 1 or uˢ⁺¹ = w^(Q+1) iff s ≡ 1 or Q mod E, or (when w^(Q+1) = 1) s ≡ −1 or −Q mod E. Rests on a splitting lemma proved here: if every element of a coset of a finite subgroup satisfies one of the two conditions and s is coprime to the subgroup's order, then one condition holds throughout. Both are [propext, Classical.choice, Quot.sound].

The answer to the Question, assembled and kernel-clean, in the abstract setting of DE13: for a commutative monoid, a multiplicatively closed S ∋ 1 of exponent two and d least positive with xᵈ ∈ S, d ∣ orderOf x and orderOf x ∣ 2d; and 2d = orderOf x iff orderOf x is even and x^(k/2) ∈ S. Composed with DE13 (eventualₚeriod_formula): the least t ≥ 1 fixing Ω = P ∪ wR pointwise up to the involution satisfies d ∣ k, k ∣ 2d, and halves exactly when n^(k/2) ∈ K_α inside Z/EZ, where K_α = 1, Q or 1, Q, −1, −Q. This is 𝓀 = k/|⟨n⟩ ∩ K_α| with the intersection of order 1 or 2.

Controls for the proof, each proved rather than asserted: (i) the splitting lemma of DE13 is FALSE without the coprimality of s and |H| — in the group of order two with s = 2 the covering hypothesis holds and neither alternative holds throughout; (ii) the dichotomy of DE14 is FALSE if S has an element of order three — with S = 1,2,4 ⊆ Z/7 and x = 3 of order 6 the least t is 2 and 6 ∤ 4, so the period is neither k nor k/2; (iii) non-vacuity: every hypothesis of the assembled theorem is discharged at the family's smallest halving case (q = 5, α = 1, n = 2, G₂ cyclic of order 3, E = π₂ = 3, k = 2), and the conclusion returns the halving branch 2d = k.

THE CRITERION OF DE5 IS A THEOREM. For every prime power q, every α ∈ 𝔽_q^× and every n ≥ 1 with αⁿ = α: 𝓀 = k / |⟨n⟩ ∩ K_α| inside (Z/π₂Z)^×, where π₂ is the largest divisor of π_q(α) coprime to n, k = ord_(π₂)(n), and K_α = 1, q (α ≠ 1) or 1, −1, q, −q (α = 1). Hence 𝓀 ∈ k, k/2 — the dichotomy the source only observes — and 𝓀 = k/2 exactly when the involution n^(k/2) lies in K_α. Proof: the double cover φ_α(u) = u + α/u of 𝔽_q by 𝔽_q^× ∪ u: u^(q+1) = α intertwines D_(n,α) with u ↦ uⁿ; the periodic set is μₐ ∪ w μ_(M₂) inside the subgroup of 𝔽_(q²)^× of order coprime to n; the fixing condition is DE13; and the exponent of the group generated by the periodic set is π₂ (case analysis on the parities of q and n, using the source's Theorem 1.1 once). The source's Proposition 4.5 and its Corollary 4.3(1) in the corrected form of DE10 are corollaries.