Three algebraic conjugates summing to zero: degrees 28 and 35 are impossible, degree 55 is not

QR(11) = 1,3,4,5,9 is the image of squaring on (Z/11)^*, equals the orbit <3> = 3ˡ: l < 5, and 3 has order exactly 5 mod 11 – so sigma: zeta₁1 -> zeta₁1³ generates Gal(Q(zeta₁1)/Q(sqrt(-11))), cyclic of degree 5.

prod_(a in QR(11)) (1 + zetaᵃ) = (1+zeta)(1+zeta³)(1+zeta⁹)(1+zeta⁵)(1+zeta⁴) = -1 for every zeta in a commutative ring with Phi₁1(zeta) = 0 (hence in Q(zeta₁1), and for any primitive 11th root of unity in a domain). Proof: the explicit factorisation (1+X)(1+X³)(1+X⁹)(1+X⁵)(1+X⁴) + 1 = Phi₁1(X) * Q(X) with Q = X¹2 - X¹0 + X⁹ + X⁸ - X⁶ + X⁴ + X³ - X² - X + 2, in Z[X].

The Hilbert-90 norm condition for the degree-55 construction: with E = Q(zeta₁1), F = Q(sqrt(-11)), sigma: zeta -> zeta³ and kappa = -(1 + zeta₁1), the relative norm N_(E/F)(kappa) = prod_(l<5) sigmaˡ(kappa) = prod_(l<5) (-(1 + zeta^(3ˡ))) equals 1. This is exactly the hypothesis under which Hilbert's Theorem 90 for the cyclic degree-5 extension E/F produces w with sigma(w)/w = kappa, i.e. w + zeta*w + sigma(w) = 0.

prod_(a in QR(7)) (1 + zeta₇ᵃ) = (1+zeta)(1+zeta²)(1+zeta⁴) = +1 whenever Phi₇(zeta) = 0. Negative control: it refutes the too-large claim 'the quadratic-residue product is -1 for every prime p >= 7'. The sign depends on p mod 8 (journal section 4.1), and 7 = 7 mod 8.

Freeness certificate for the degree-55 orbit: (1 + zeta₁1)¹1!= -1 in any commutative ring of characteristic zero with Phi₁1(zeta) = 0, equivalently kappa = -(1 + zeta₁1) satisfies kappa¹1!= 1, so kappa is not an 11th root of unity. Hence Stab_G(w) = 1 for G = mu₁1: <sigma> of order 55 and the orbit G.w has full size 55. Bezout certificate: U(X)*((1+X)¹1 + 1) + V(X)*Phi₁1(X) = 15709 in Z[X].

Non-vacuity of the hypothesis 'Phi₁1(zeta) = 0': zeta = 64 in Z/67 satisfies 1 + zeta +... + zeta¹0 = 0, zeta!= 1 and zeta¹1 = 1, and the identity of T2 holds there (the product is 66 = -1).

Negative controls for the identity: over Z/67 at zeta = 64 the product over the wrong exponent set 1,2,3,4,5 is 58, neither -1 nor 1; and the QR(11) product is not +1. So T2 is a statement about the quadratic residues, not about any five exponents, and its sign is load-bearing (Hilbert 90 needs (-1)⁵ * (-1) = 1).

The p-group step of the source's Theorem t1: if a p-group H acts transitively on a finite nonempty set Omega and W is a nonzero H-invariant Fₚ-subspace of Fₚ^Omega, then the all-ones vector lies in W.

Theorem t1, mod-p form: if a p-group acts transitively on Omega and A is a proper invariant Fₚ-subspace of Fₚ^Omega, then every a in A has zero coordinate sum. Includes the duality step (a proper subspace has a nonzero dot-product-orthogonal vector, and orthogonality is preserved by the coordinate action).

The source's Lemma l1, prime-power case: a Sylow p-subgroup of a finite group acting transitively on a set of size pᵐ acts transitively itself.

Theorem t1, integer form: for an integer vector c on Omega whose mod-p reduction lies in a proper invariant subspace of Fₚ^Omega under a transitive p-group, p divides sumₓ cₓ. With Omega the roots of an irreducible f of degree pᵐ, the Galois group's Sylow p-subgroup (transitive by T10) and A the mod-p reduction of the relation lattice, this is the source's Theorem t1 verbatim.

Theorem p2, prime-power half: with the same hypotheses, if A contains the indicator vector of a triple (i,j,k) then p = 3; and the signed relation alphaᵢ + alphaⱼ - alphaₖ = 0 is impossible for every prime p, because its coefficient sum is 1. At p = 2, m = 4 the first statement is the source's title result, d!= 16.

Controls for the prime-power theorems: each of the three hypotheses is necessary – S₃ on Fin 3 is transitive and preserves the proper line spanned by the all-ones vector, whose coordinate sum is 1!= 0 in ZMod 2 (so 'p-group' is needed); the trivial group on Fin 2 preserves the proper line spanned by e₀ (so 'transitive' is needed); A = top contains e₀ (so 'proper' is needed). The hypotheses are jointly satisfiable with a nontrivial A (Perm (Fin 2) on Fin 2 and the all-ones line). The conclusion p = 3 is sharp: (3: ZMod q) = 0 iff q = 3.

d = 28: there is NO algebraic number of degree 28 three of whose conjugates sum to zero. Proof: the unconditional direction of the Girstmair reduction (T19) turns such a number into a transitive group G of degree 28 with a 3-subset orbit whose Q[G]-module Q[G].e_T contains no eₐ - e_b; running that test on all 1854 transitive groups of degree 28 (GAP + TransGrp 2.0.6, exact rationals) over every 3-subset orbit returns no passing pair.

d = 35: same conclusion, by the same procedure over all 407 transitive groups of degree 35.

d = 55: there IS an irreducible f in Q[t] of degree 55 with three distinct roots summing to zero. Construction: E = Q(zeta₁1), sigma: zeta -> zeta³ of order 5, F = E^<sigma> = Q(sqrt(-11)), kappa = -(1 + zeta); N_(E/F)(kappa) = 1 (row T3) so Hilbert 90 gives w with sigma(w)/w = kappa, i.e. w + zeta*w + sigma(w) = 0; kappa is not an 11th root of unity (row T5) so the orbit of w under G = mu₁1: <sigma> = C₁1: C₅ has size 55; G is solvable of odd order hence a Galois group over Q (Scholz-Reichardt), and the normal basis theorem transfers the relation to an algebraic number of degree 55. Consequently d = 55s is attainable for every s >= 1, in particular d = 110.

An infinite family generalising d = 55: for every prime p = 11 (mod 24) with (p-1)/2 prime, there is an algebraic number of degree d = p(p-1)/2 with three conjugates summing to zero, and d is divisible by neither 3 nor 20. First cases p = 11, 59, 83, 107, 179 giving d = 55, 1711, 3403, 5671, 15931. d = 55 is the only member below 100.

Sign law: for a prime p >= 7 and zeta a primitive p-th root of unity, prod_(a in QR(p)) (1 + zetaᵃ) = +1 if p = 1 or 7 (mod 8), = -1 if p = 3 (mod 8), and is not a rational integer if p = 5 (mod 8). Proof in the journal (section 4.1) via 1 + zetaᵃ = (1 - zeta²ᵃ)/(1 - zetaᵃ) and the unit structure of the quadratic subfield; verified for every prime 5 <= p <= 200 in PARI.

The per-degree decision procedure: an algebraic number of degree d with three DISTINCT conjugates summing to zero exists if and only if there are a finite group G occurring as a Galois group over Q, a transitive action of G on a d-set Omega, and a 3-subset T with the Q[G]-submodule Q[G].e_T containing no vector eₐ - e_b (a!= b). The forward direction needs no hypothesis; the converse uses the normal basis theorem and realisability of G over Q. Executed in GAP by spinning e_T under the generators with an RREF basis.

Compute-first-values gate, reproduced in this session: the sixteen open degrees are exactly [2,100] minus multiples of 3, multiples of 20, prime powers and 2pᵐ (p >= 5); Stong's t²0 + 4*5⁹ t¹0 + 16*5¹5 is irreducible and has exactly 20 zero-sum root triples, the first being roots (1,7,10); the decision procedure returns 0 passing pairs over all 1954 transitive groups of degree 16 (the source's headline) and 7 over the 1117 of degree 20 (T20#5, 9, 15, 30, 35, 36, 65, every one with dim N = 16), and reproduces the expected answer at degrees 3, 5, 6, 9, 10, 12, 14, 15, 21, 22. Cost: 80.0 CPU-s for the fourteen-degree GAP sweep, under 1 CPU-min for PARI, under 1 GB peak.