The affine Bernstein basis on the q-quadratic lattice admits no first-order lattice ladder beyond degree two

Conjecture conj:ladder of arXiv:2608.04802v1, settled and sharpened: for every n >= 3 (the source asks n >= 4), every 0 <= k <= n and every q in (0,1), the first-order lattice ladder sigma * Dₓ Bⁿₖ = tau * Sₓ Bⁿₖ with deg sigma <= 2 and deg tau <= 1 forces sigma = tau = 0. Proved uniformly in n from the norm identity QB4 and the coprimality QB5, computing no determinant. Both the mu form (ladderₜrivial) and the q form (conjₗadder_q) are bound here.

The conjugate-pair representation that justifies the definitions: for every real x and every d with d² = (mu²-1)(x²-1), Bⁿₖ(mu*x + d) = (Sₓ Bⁿₖ)(x) + d (Dₓ Bⁿₖ)(x), and likewise with -d. This pins Sp and Dp as the two components of the half-step value in the quadratic extension R[x][d]/(d² - (mu²-1)(x²-1)).

The lattice bridge: on the q-quadratic lattice x(s) = (qˢ + q⁻s)/2 with Q = q^(1/2), mu = (Q+Q⁻1)/2, the two half-step abscissae satisfy x(s+1/2) + x(s-1/2) = 2 mu x(s) and ((x(s+1/2)-x(s-1/2))/2)² = (mu²-1)(x(s)²-1); consequently the polynomials Sp, Dp evaluated at x(s) are exactly the source's Sₓ Bⁿₖ and Dₓ Bⁿₖ.

The norm identity: (Sₓ Bⁿₖ)² - (mu²-1)(x²-1) (Dₓ Bⁿₖ)² = (C(n,k)/2ⁿ)² (x+mu)²ᵏ (x-mu)^(2(n-k)) for every n and every k <= n. Equivalently, the product of the two half-step values of the affine Bernstein element factors completely, with the only roots x = +-mu.

Sₓ Bⁿₖ and Dₓ Bⁿₖ are coprime in R[x], for every n, every k <= n and every q in (0,1) – the rank statement the source says is unavailable for n >= 4. Proved by a constructive Bezout chain: (Sₓ Bⁿₖ)(mu) and (Sₓ Bⁿₖ)(-mu) are nonzero, so S is coprime to (x-mu) and (x+mu), hence to the right-hand side of the norm identity, hence to D.

Sharpness of QB1 in n: for n <= 2, every k <= n and every q in (0,1), (sigma, tau) = (Sₓ Bⁿₖ, Dₓ Bⁿₖ) is a nonzero solution of eq:ladder with deg sigma <= 2 and deg tau <= 1. With QB1 this gives the exact dichotomy: eq:ladder has a nonzero solution if and only if n <= 2.

mu = (q^(1/2) + q^(-1/2))/2 > 1 for every q in (0,1), so the q form of the conjecture (conjₗadder_q) follows from the mu form.

Negative controls. (i) Non-vacuity: Dₓ Bⁿₖ is not the zero polynomial for n >= 3 and mu > 1, so eq:ladder is a genuine constraint. (ii) Too-small control: the degree bounds deg sigma <= 2, deg tau <= 1 cannot be dropped – without them (Sₓ Bⁿₖ, Dₓ Bⁿₖ) is a nonzero solution for every n. (iii) The hypothesis on mu is not decoration: at mu = 1 (that is q = 1) the classical ladder (1-x²) Dₓ B³₀ = (-3-3x) Sₓ B³₀ is a nonzero solution; at mu = -1 so is (1-x², -3+3x); at mu = 0 one has Sₓ B³₁ = -Dₓ B³₁, so (1,-1) solves.

Reproduction of the source's four displayed degree-three polynomials, from the definitions, in the kernel: 8 Dₓ B³₀ = (mu²-4) + 6 mu x - (4 mu²-1) x²; 8 Sₓ B³₀ = (4-3mu²) + 3mu(mu²-2)x + 3(2mu²-1)x² - mu(4mu²-3)x³; (8/3) Dₓ B³₁ = -mu² - 2 mu x + (4mu²-1)x²; (8/3) Sₓ B³₁ = mu² - mu(3mu²-2)x - (2mu²-1)x² + mu(4mu²-3)x³.

The source's eq:ladderdet reproduced exactly: with 8 M_(3,0) and (8/3) M_(3,1) the coefficient matrices the source displays (columns = the coefficient vectors of Dₓ B³ₖ, x Dₓ B³ₖ, x² Dₓ B³ₖ, -Sₓ B³ₖ, -x Sₓ B³ₖ), det(8 M_(3,0)) = 64 (mu²-1)⁶ and det((8/3) M_(3,1)) = -64 mu⁸ (mu²-1)², i.e. det M_(3,0) = (mu²-1)⁶/512 and det M_(3,1) = -243 mu⁸ (mu²-1)²/512 as printed, and det M_(3,2) = -det M_(3,1), det M_(3,3) = -det M_(3,0).

The exact degenerate locus of the ladder. For every n >= 3, every k <= n and every REAL mu with mu!= 0 and mu²!= 1, eq:ladder admits only the trivial solution – equivalently the (n+2) x 5 coefficient matrix M_(n,k) has full column rank 5. Together with QB8(iii), which exhibits a nonzero degree-three solution at each of mu = 0, 1, -1, the rank of M_(n,k) drops exactly on mu = 0, 1, -1. q in (0,1) is the special case mu > 1.

Computed 5th determinantal divisor of the ladder matrix M_(n,k) (the gcd of all C(n+2,5) maximal minors, whose roots are the rank-drop points), for 3 <= n <= 7 and every k: always c * mu^(a(n,k)) (mu²-1)^(b(n,k)) with c a nonzero rational, b(n,k) = 6 for k in 0,n and 2 otherwise, and a(n,k) = (0,8,8,0) for n=3, (0,5,9,5,0) for n=4, (0,0,8,8,0,0) for n=5, (0,0,5,9,5,0,0) for n=6, (0,0,0,8,8,0,0,0) for n=7.