Quartic Centroid
Conjecture

Position Moves. Residue Does Not.

For the 4-term arithmetic product family (x+k)(x+k+2)(x+k+4)(x+k+6) = (n−1)², the solution decomposes into two fully independent components: a centroid that shifts with k, and a residue that is fixed regardless of where the sequence starts. The two components are decoupled — a direct instance of MPRC's core architecture separating movement from invariant structure.

The Three Invariants

Invariant I — Centroid. x_base = −(k+3). The centroid operator always inverts the parity of k. If k is odd, x_base is even. If k is even, x_base is odd. Adding 3 (odd) to any k always flips parity — structural, not coincidental.

Invariant II — Residue. x_res = ±√10 = ±√(1²+3²). The residue is fixed and k-independent. It encodes the geometry of the 4-step family: the four terms sit at centroid ±1 and centroid ±3. The residue is the Euclidean distance over these offsets. It does not change regardless of where the sequence starts.

Invariant III — Scaling Law. Δx_base = −Δk. Δx_res = 0. Position moves. Residue does not. They are fully decoupled.

Shape Geometry, Not Arbitrary Choice

The RHS is not arbitrary. It is the geometric area of the shape defined by the family:

90 Seconds vs. 10 Minutes

Solve (x+2)(x+4)(x+6)(x+8) = 105.

Standard method: Expand to x⁴ + 20x³ + 143x² + 430x + 279 = 0, then apply rational root theorem testing ±1, ±3, ±9, ±31… Minimum 10 minutes. No geometric insight.

Conjecture method: 4 terms, step=2. Centroid = (2+4+6+8)/4 = 5. x_base = −5. Clean residue check: 16 + 105 = 121 = 11². Perfect square — clean roots guaranteed before solving. Let t = x+5: (t²−9)(t²−1) = 105. Then u² − 10u − 96 = 0 → (u−16)(u+6) = 0 → u = 16 → t = ±4 → x = −1 or x = −9.

Residue² = 5 + √(16+N)

Clean residues occur exactly when 16+N is a perfect square. N=9 gives the first clean surd. N=105 gives the first clean integer residue beyond N=0.

Negative k and Step ≠ 2

Test 1 — Negative k (k = −1). Prediction: x = −2 ± √10. Equation: (x−1)(x+1)(x+3)(x+5) = 9. Let t = x+2: (t²−9)(t²−1) = 9 → u² − 10u = 0 → u = 10 → x = −2 ± √10 ✓. Parity: k = −1 odd → x_base = −2 even. Inversion holds. Scaling: Δk = −2 → Δx_base = +2. CONJECTURE SURVIVES.

Test 2 — Step = 3 (k = 0). Equation: x(x+3)(x+6)(x+9) = 9. Centroid = 18/4 = 4.5 → x_base = −4.5. Offsets: ±1.5, ±4.5. Residue² = 1.5² + 4.5² = 22.5. Residue = 1.5√10 = (step/2)·√10. CONJECTURE SURVIVES.

√10 Survives All Step Changes

Step is a scaling factor only. The geometric invariant √10 = √(1²+3²) is the Euclidean distance over the offsets {±1, ±3} at unit step — it survives all step changes.

f(n) = 3n — Open Connection

For any regular n-gon, connecting all vertices to centre C and bisecting each corner angle produces exactly 3n angles total: n central angles at C, n bisected corner angles, and 2n small angles from each bisection.