A Geometric Framework and Complete Proof of the Beal Conjecture

We present a complete proof that the equation Ax+By=CzA^x+B^y=C^z 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))μ(A,B,C)=log∏_p pmin(v_p(A),v_p(B),v_p(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 μμ μ, the minimum harmonic-mode bound μ2≥3μ1μ_2≥ 3μ_1 μ2≥3μ1, equivalence of Beal with μ≥log⁡2μ≥log 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=A^x R=Ax, S=ByS=B^y S=By, the condition j(E)=jCMj(E)=jCM j(E)=jCM is equivalent to the cubic 256(u−1)3=jCM(u+2)256(u-1)^3=jCM(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)c_4=16(a^2+ab+b^2) c4=16(a2+ab+b2) satisfies v2(c4)=4v_2(c_4)=4 v2(c4)=4, giving f2≤4f_2≤ 4 f2≤4 and N0≤16N_0≤ 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≤16N_0≤ 16 N0≤16. Since dim⁡S2(Γ0(N0))=0 S_2(Γ_0(N_0))=0 dimS2(Γ0(N0))=0 for every N0∈{1,2,4,8,16}N_0∈\{1,2,4,8,16\} N0∈{1,2,4,8,16}, no such newform exists. The geometric interpretation: the coprime zone {μ1=0}\{μ_1=0\} {μ1=0} is surrounded by infinite acoustic impedance Z=μ2/μ12≥3/μ1→∞Z=μ_2/μ_1^2≥ 3/μ_1→∞ Z=μ2/μ12≥3/μ1→∞, corresponding precisely to the modular-forms contradiction.