írj egy appot python es html The backend must always be impl...

Prompt

írj egy appot python es html The backend must always be implemented in a single file, and the frontend must always be contained in a single index.html file, including all subpages. azt akarom hogy ilyen es ehhez hasonlo kepleteket: $$ \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 \\[12pt] \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) \\[12pt] \forall p \ge 5 : \begin{cases} \mathbf{P}^{(p)}(x, y) = -v(x, y) \sum_{k=0}^{p-2} \Big(-u(x, y)v(x, y)\Big)^k \in \mathbb{F}_p[x, y] \\ \mathbf{Q}^{(p)}(x, y) = u(x, y)\Big(1 + u(x, y)v(x, y)\Big) \in \mathbb{F}_p[x, y] \end{cases} \\[14pt] \mathbf{F}_{\mathcal{U}} = \Big[ \big(\mathbf{P}^{(p)}, \mathbf{Q}^{(p)}\big) \Big]_{p \in \mathcal{U}} : \prod_{p \in \mathcal{U}} \overline{\mathbb{F}}_p^2 \to \prod_{p \in \mathcal{U}} \overline{\mathbb{F}}_p^2 \quad \left(\prod_{p \in \mathcal{U}} \overline{\mathbb{F}}_p \cong \mathbb{C}\right) \\[12pt] \mathcal{S} = \left\{ (x, y, w) \in \mathbb{C}^3 \;\middle|\; y = w^2 + \dfrac{32}{9} x^{15} w^9 \right\} \cong \mathbb{A}^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} \\[16pt] \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 = \mathbf{1} \quad \text{mindenütt a reguláris szimplektikus fázistéren } \mathbb{C}^2 \setminus \{1 + uv = 0\} \\[10pt] \wedge\quad & \forall p \ge 5 : \quad \det J_{(x, y)}\big(\mathbf{P}^{(p)}, \mathbf{Q}^{(p)}\big) = 1 + (-1)^{p-2} p \cdot (uv)^{p-1} \equiv \mathbf{1} \pmod p \quad \text{az affin polinomgyűrűben } \mathbb{F}_p[x, y] \\[10pt] \wedge\quad & \det J\big(\mathbf{F}_{\mathcal{U}}\big) = \Big[ \det J\big(\mathbf{P}^{(p)}, \mathbf{Q}^{(p)}\big) \Big]_{\mathcal{U}} \equiv \mathbf{1} \quad \text{a karakterisztika-0 algebrailag zárt test felett } \prod_{p \in \mathcal{U}} \overline{\mathbb{F}}_p \cong \mathbb{C} \\[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) \\[12pt] \wedge\quad & \mathbf{F}_{\mathcal{S}}(0, 1) = \mathbf{F}_{\mathcal{S}}(0, -1) = (0, 1) \implies \neg\operatorname{Inj}(\mathbf{F}_{\mathcal{S}}) \\[10pt] \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 \\[10pt] \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 \ge 5 \\[10pt] \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\} \\[10pt] \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] = (1\,2)(3\,4) \ne \operatorname{id} \implies \text{Nem-abeli étale monodrómia} \\[12pt] \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 + u(x, y)v(x, y)} = \Omega_{\log} \\[12pt] \wedge\quad & \neg\operatorname{Proper}(\mathbf{F}) \;\wedge\; \neg\operatorname{Bir}(\mathbf{P}_{\mathcal{S}}, \mathbf{Q}_{\mathcal{S}}) \;\wedge\; \neg\operatorname{Inv}(\mathbf{F}_{\mathcal{S}}) \end{aligned} $$ frontenden be illesztek DE BARMILYEN KEPLETTEL MUKODJON es a backend a TELJES KEPLETET az utolso betu/szamig mindent le futtar es vegul ki irja h ervesnyes counterexample e a jacobi sejtesre 2d ben. de ne ugy csinald hogy SZANDEKOSAN csak azt irja ji hogy nem csak azert mert te egy papagaj vagy es csak annyit tudsz szajkozni hogy remains open remain open es MINDEN FELE KEPLETTEL MULODJON Send csak a main.pyt ird a frontend nekem mar kesz van