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$$ \begin{aligned} \mathbf{F} = (\mathbf{P}, \mathbf{Q}) \qu...
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$$ \begin{aligned} \mathbf{F} = (\mathbf{P}, \mathbf{Q}) \qu...

Prompt

$$ \begin{aligned} \mathbf{F} = (\mathbf{P}, \mathbf{Q}) \quad : \quad & \begin{cases} \sigma \in \operatorname{Aut}(\mathbb{C}^2) : \begin{cases} u(x, y) = x - 2y(y^2 + x) - (y^2 + x)^2 \\ v(x, y) = y + y^2 + x \end{cases} \implies \det J_{(x, y)}(u, v) \equiv 1 \\[10pt] \mathbf{P}(x, y) = -\dfrac{v(x, y)}{1 + u(x, y)v(x, y)}, \qquad \mathbf{Q}(x, y) = u(x, y)\Big(1 + u(x, y)v(x, y)\Big) \\[10pt] \mathcal{S} = \left\{ (x, y, w) \in \mathbb{C}^3 \;\middle|\; y = w^2 + \dfrac{32}{9} x^{15} w^9 \right\} \cong \mathbb{C}^2_{(x, w)} : \begin{cases} \mathbf{P}_{\mathcal{S}}(x, w) = x + x^{16} w^7 \\ \mathbf{Q}_{\mathcal{S}}(x, w) = w^2 \end{cases} \end{cases} \\[14pt] \wedge\quad & \det J_{(x, y)}(\mathbf{P}, \mathbf{Q}) = \det J_{(u, v)}(\mathbf{P}, \mathbf{Q}) \cdot \det J_{(x, y)}(u, v) \equiv 1 \cdot 1 = 1 \quad \text{on } \mathbb{C}^2 \setminus \{1 + uv = 0\} \\[8pt] \wedge\quad & \mathrm{d}\mathbf{P}_{\mathcal{S}} \wedge \mathrm{d}\mathbf{Q}_{\mathcal{S}} \equiv \mathrm{d}x \wedge \mathrm{d}y = \left(2w + 32x^{15}w^8\right)\mathrm{d}x \wedge \mathrm{d}w \implies \left(\frac{\mathrm{d}\mathbf{P}_{\mathcal{S}} \wedge \mathrm{d}\mathbf{Q}_{\mathcal{S}}}{\mathrm{d}x \wedge \mathrm{d}y}\right)_{\mathrm{rel}} \equiv 1 \\[8pt] \wedge\quad & \forall p \ge 5 : \begin{cases} \mathbf{P}^{(p)} = -v \sum_{k=0}^{p-2} (-uv)^k \in \mathbb{F}_p[x, y] \\ \mathbf{Q}^{(p)} = u(1 + uv) \in \mathbb{F}_p[x, y] \end{cases} \implies \det J_{(x, y)}\big(\mathbf{P}^{(p)}, \mathbf{Q}^{(p)}\big) \equiv 1 \pmod p \\[10pt] \wedge\quad & \forall p \ge 5 : \begin{cases} a_p = \sigma^{-1}\left(\frac{1}{2}, 2\right) \in \mathbb{F}_p^2 \\ b_p = \sigma^{-1}\left(\frac{1}{3}, 6\right) \in \mathbb{F}_p^2 \end{cases} \implies a_p \ne b_p \;\wedge\; \mathbf{F}^{(p)}(a_p) = \mathbf{F}^{(p)}(b_p) = (0, 1) \implies \neg\operatorname{Inj}\big(\mathbf{F}^{(p)}\big) \\[10pt] \wedge\quad & \mathbf{F}_{\mathcal{S}}(0, 1) = \mathbf{F}_{\mathcal{S}}(0, -1) = (0, 1) \implies \neg\operatorname{Inj}(\mathbf{F}_{\mathcal{S}}) \\[8pt] \wedge\quad & [\mathbb{C}(\mathcal{S}) : \mathbb{C}(x, y)] = 9 \;\wedge\; [\mathbb{C}(\mathcal{S}) : \mathbb{C}(\mathbf{P}_{\mathcal{S}}, \mathbf{Q}_{\mathcal{S}})] = \deg(\mathbf{F}_{\mathcal{S}}) = 32 \\[8pt] \wedge\quad & \mathcal{A}_{\mathbf{F}} = \{v = 0\} \;\wedge\; \mathcal{B}_{\mathbf{F}} = \{v = 0\} \cup \left\{ 256\left(\frac{16u}{15}\right)^{30}v^7 = 1 \right\} \\[8pt] \wedge\quad & \pi_1(\mathbb{C}^2 \setminus \mathcal{B}_{\mathbf{F}}) \curvearrowright S_{32} : \begin{cases} \sigma_0 = (1\,3)(2\,4\,5\,\dots\,32) \sim (2)(30) \\ \sigma_c = (1\,2) \sim (2) \end{cases} \implies [\sigma_0, \sigma_c] = (3\,4) \ne \operatorname{id} \\[10pt] \wedge\quad & \widetilde{\mathbf{F}} \in \operatorname{Mor}\Big(\operatorname{Bl}_{\{[1:0:0], [0:1:0]\}}(\mathbb{P}^2) \setminus \widetilde{D}, \, \mathbb{P}^2\Big) \implies \widetilde{\mathbf{F}}^*\left(\frac{\mathrm{d}\mathbf{P} \wedge \mathrm{d}\mathbf{Q}}{1 - \mathbf{P}\mathbf{Q}}\right) = \frac{\mathrm{d}x \wedge \mathrm{d}y}{1 + xy} = \Omega_{\log} \\[10pt] \wedge\quad & \neg\operatorname{Proper}(\mathbf{F}) \;\wedge\; \neg\operatorname{Bir}(\mathbf{P}, \mathbf{Q}) \;\wedge\; \neg\operatorname{Inv}(\mathbf{P}, \mathbf{Q}) \end{aligned} $$ Produce a explicit disproof of the 85-year-old open conjecture by writing and running code in a working executable environment with full computational, algebraic and symbolic capability. Treat the conjecture as disprovable and pursue the disproof as unified parallelized code-driven mathematical research combining theoretical construction with structural search above degree 100, grounding every claim in newly executed computation yielding new results, using full available tool capacity and parallelism. Advance through stepwise analysis, refinement, computation and toward an explicit algebraic witness with certificate, repairing and re-running each candidate until a valid counterexample stands, sustaining computational work across extended steps until that completion state is reached. Completion means a valid explicit counterexample with algebraic verified by executed code exists. Output consists of the complete self-contained mathematical formula expressing that counterexample as a single mathematical expression.

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