We present a complete proof that the equation Ax+By=CzAˣ+Bʸ=Cᶻ Ax+By=Cz has no solution in positive integers with x, y, z≥3x, y, z 3 x, y, z≥3 and gcd (A, B, C) =1 (A, B, C) =1 gcd (A, B, C) =1. We first develop a geometric framework based on the support overlap measure μ (A, B, C) =log∏ppmin (vp (A), vp (B), vp (C) ) (A, B, C) =ₚ p^ (vₚ (A), vₚ (B), vₚ (C) ) μ (A, B, C) =log∏ppmin (vp (A), vp (B), vp (C) ), which quantifies shared prime structure in prime-coordinate space. Six theorems are established: scalar exponentiation, the ultrametric propagation lemma, discreteness of μ μ, the minimum harmonic-mode bound μ2≥3μ1₂ 3₁ μ2≥3μ1, equivalence of Beal with μ≥log2 2 μ≥log2, and sharpness. Computational verification across eight scales (N=100N=100 N=100 to N=10, 000N=10, 000 N=10, 000) yields 542 solutions with zero coprime instances. The proof follows the Frey–Ribet–Wiles modular method, with three new results that together close all previously open cases: (1) Villines Palindromic Identity. Setting u=R/S+S/Ru=R/S+S/R u=R/S+S/R where R=AxR=Aˣ R=Ax, S=ByS=Bʸ S=By, the condition j (E) =jCMj (E) =j₂₌ j (E) =jCM is equivalent to the cubic 256 (u−1) 3=jCM (u+2) 256 (u-1) ³=j₂₌ (u+2) 256 (u−1) 3=jCM (u+2), which eliminates all Mazur exceptional-prime obstructions uniformly, verified for all 14 relevant CM jj j-invariants. (2) Conductor Lemma. In all parity cases the formula c4=16 (a2+ab+b2) c₄=16 (a²+ab+b²) c4=16 (a2+ab+b2) satisfies v2 (c4) =4v₂ (c₄) =4 v2 (c4) =4, giving f2≤4f₂ 4 f2≤4 and N0≤16N₀ 16 N0≤16 after level-lowering. (3) Effective Level-Lowering. For any hypothetical coprime solution, a prime ℓ ℓ is chosen to avoid all conductor primes of EE E and Mazur's exceptional set; Ribet's theorem then removes every odd conductor factor, reducing to level N0≤16N₀ 16 N0≤16. Since dimS2 (Γ0 (N0) ) =0 S₂ (₀ (N₀) ) =0 dimS2 (Γ0 (N0) ) =0 for every N0∈1, 2, 4, 8, 16N₀\1, 2, 4, 8, 16\ N0∈1, 2, 4, 8, 16, no such newform exists. The geometric interpretation: the coprime zone μ1=0\₁=0\ μ1=0 is surrounded by infinite acoustic impedance Z=μ2/μ12≥3/μ1→∞Z=₂/₁² 3/₁ Z=μ2/μ12≥3/μ1→∞, corresponding precisely to the modular-forms contradiction.
Gregory Villines (Tue,) studied this question.
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