EXACT CLAIM

Let p and l be distinct odd primes satisfying p≡2 (mod 3), l≡7 (mod 24), and (p/l)=-1, where the parentheses denote the Legendre symbol. For every a,b≥1, set q=p^a l^b. There are no integers n≥1 and K>1 and no rank-K orthogonal projector P on (C^q)^{⊗n} satisfying both equations (1) and (3) in the problem, for the literal cyclic Weyl errors of weight at most two. In particular this excludes every q=5^a7^b and every q=17^a7^b. No stabilizer or prime-power assumption on K is made. As an ingredient, for any integer q≥2, equations (1) and (3) force the degree-two polynomial L₂(x)=Σ_{j=0}²[z^j](1+(q²−1)z)^{n−x}(1−z)^x to have two distinct integer roots in [5,n]. This is a partial nonexistence theorem, not a solution for all non-prime-power local dimensions.

FULL FROZEN PROOF

1. An alphabet-independent quantum Lloyd argument.
Write D=q^n, Q=q², and W_g for the tensor-product cyclic Weyl operators, indexed by g in (Z/qZ)^{2n}. These form an orthogonal operator basis: tr(W_g†W_h)=Dδ_{g,h}. Every nonidentity Weyl operator of weight at most four is, up to phase, E†F for distinct errors E,F of weight at most two: partition its nonidentity sites into two sets of size at most two, and choose the local inverse operators on the first set. Thus purity implies PW_gP=0 at weights 1 through 4. Consequently tr(W_g†P)=0 at these weights, since tr(W_g†P)=tr(PW_g†P).

Define T_j(A)=Σ_{wt(g)=j}W_g A W_g†. Conjugation acts diagonally on the Weyl basis. At a single site the sum of its conjugation characters over all q² labels is q² for the identity label and zero for every nonidentity label. This holds for composite q as well: each sum factors into geometric sums of powers of the primitive qth root of unity, and a nonzero label gives at least one vanishing sum. After deleting the identity conjugator, the corresponding sums are Q−1 and −1. Therefore, for wt(h)=w,
T_j(W_h)=K_j(w)W_h,
where K_j(w)=[z^j](1+(Q−1)z)^{n−w}(1−z)^w.

The subspaces E(C), for wt(E)≤2, are mutually orthogonal by (1), and perfection says their dimensions sum to D. Hence Σ_{j=0}²T_j(P)=I. Expanding P in the Weyl basis shows that every nonidentity weight w appearing with a nonzero coefficient in P satisfies L₂(w)=0. Such weights are at least five by purity.

There must be at least two distinct such weights. Indeed, n=1 is impossible because perfection would give Kq²=q. Thus n≥2, and a weight-two Weyl operator F exists. Suppose that the nonidentity support weights of P comprise at most one value. Choose a polynomial β of degree at most one with β(0)=D/K and vanishing on that support. If the support is empty, choose β constant. As K₀(x)=1 and K₁(x)=(Q−1)n−Qx span the polynomials of degree at most one, write β=β₀K₀+β₁K₁. The diagonalization just proved yields β₀T₀(P)+β₁T₁(P)=I. Right-multiply by FP. Every summand on the left is a multiple of EP E†FP with wt(E)≤1. Here E†F is a nonidentity Weyl operator up to phase and has weight at most three. Purity makes each such summand zero. The right side FP is nonzero, a contradiction. Since L₂ is quadratic with leading coefficient Q²/2, its roots are therefore two distinct integers r,s with 5≤r<s≤n.

2. Exact root arithmetic for the claimed family.
Now assume q=p^a l^b as in the statement. In particular q is odd and coprime to 3. Put A=Q−1 and V=1+nA+n(n−1)A²/2. Direct expansion gives
2L₂(x)=Q²x²−Q(2An−Q+4)x+2V.
Thus
S:=r+s=(2An−Q+4)/Q,
R:=rs=2V/Q²,
Δ:=(s−r)²=1+4A(n−2)/Q².
Because Δ is an integer and gcd(Q²,4A)=1 for odd Q, we have Q² | n−2. Write n=2+Q²u. The root bounds imply n≥6, so the integer u is positive. Substitution gives
S=3+2QAu,
Δ=1+4Au,
R=2+(3Q−1)Au+Q²A²u².
These are exact identities.

Perfection implies V divides q^n because K is an integer. Since Q²R=2V and R is an integer, every prime divisor of R belongs to {2,p,l}. As the square of an odd integer, Q is 1 modulo 8. The formula for R consequently gives R≡2 (mod 8). In particular exactly one factor of 2 divides R.

Neither p nor l divides both r and s. For either prime h in {p,l}, Q≡0 (mod h), so S≡3 (mod h). Both primes differ from 3 by the hypotheses. This also handles the possible exceptional-looking prime p=5: the sum is 3, not 0, modulo 5, so it cannot divide both roots.

Each root has an odd prime divisor. Otherwise it would be a power of 2, and because the product contains exactly one factor of 2, that root would be at most 2, contrary to its lower bound 5. There are only two available odd primes, and neither can occur in both roots. Therefore both primes occur, one in each root. Up to exchanging r and s the only possibilities are
{r,s}={p^i,2l^j} or {r,s}={2p^i,l^j},
with i,j positive integers.

3. The modular contradiction.
Modulo l, the root divisible by l is zero and S≡3. Thus the other root is 3 modulo l. In the first case p^i≡3 (mod l); in the second case 2p^i≡3 (mod l).

For a prime l≡7 (mod 24), the supplementary law for 2 gives (2/l)=1, since l≡7 (mod 8). Quadratic reciprocity gives (3/l)=-1, since l≡7 (mod 12). These formulas have no exceptional prime here: l≥7. Taking Legendre symbols in either root congruence and using (p/l)=-1 therefore yields (-1)^i=-1. Hence i is odd.

But p≡−1 (mod 3) and l≡1 (mod 3). For odd i, the two roots in the first case are both −1 modulo 3; in the second case they are both 1 modulo 3. Thus Δ=(s−r)²≡0 (mod 3). On the other hand q is coprime to 3, so Q≡1 (mod 3), and the identity Δ=1+4(Q−1)u gives Δ≡1 (mod 3). This contradiction proves the theorem.

For the examples, (5/7)=-1 and (17/7)=(3/7)=-1, while 5 and 17 are 2 modulo 3 and 7 is 7 modulo 24. The exclusion p=2 is essential to the stated hypotheses: the odd-Q divisibility and modulo-8 arguments were not asserted for it.

4. Scope and verification.
The argument is analytic; finite checking is not used to prove impossibility. The accompanying standalone code verifies representative prime-pair conditions, the precise arithmetic identities and residue obstructions, and rejected controls. It uses no external files or numerical approximations. Relative to the supplied source summaries, Bennett's largest-prime-at-most-13 classical result already covers the classical 5,7 family. No new classical theorem is claimed here. The quantum conclusion above is established directly, including its Lloyd premise, rather than inferred from a classical code's existence. The 17,7 family lies outside that stated largest-prime-at-most-13 result, and neither family is covered by the supplied theorem concerning alphabets with a factor 2. No stronger claim about overlap with every argument in Bennett or the latest preprint is made.

REMAINING GAP

The existence question for arbitrary non-prime-power q remains open outside the proved and previously supplied exclusion families. No code is constructed and no universal nonexistence theorem is proved. A complete audit of overlap with all arguments in Bennett and the latest preprint is not supplied.