## 2. Results/Conclusions - **explicit counterexample:** $...

Prompt

## 2. Results/Conclusions - **explicit counterexample:** $$ \begin{aligned} F = (P, Q): \mathbb{C}^2 \to \mathbb{C}^2 \quad : \quad & \begin{cases} P(x, y) = x - \frac{35}{24} x^{12} y^7 + y^7 \sum_{s=0}^{11} a_s x^s + x^{16} y^{16} & (\deg P = 32) \\[8pt] Q(x, y) = y + \frac{96}{805} x^4 y^{10} + \left( \frac{7}{2} x^{41} y^{40} + \frac{35}{8} x^{26} y^{24} + \frac{35}{16} x^{11} y^8 - \frac{245}{48} x^{52} y^{47} \right) \\[6pt] \qquad\quad + \sum_{m=0}^3 y^{16m+8} \left[ \binom{7/2}{3-m} x^{11+15m} \left(1 + \frac{16}{x}\right)_{\le 11+15m}^{\frac{2m+1}{2}} + K_m(x) \right] \\[6pt] \qquad\quad + \sum_{j=1}^7 y^{\nu_j} M^{(j)}(x) + x^{56} y^{56} & (\deg Q = 112) \end{cases} \\ \wedge\quad & \gcd(\deg P, \deg Q) = 16 \;\wedge\; \neg\operatorname{Prime}(\gcd(\deg P, \deg Q)) \;\wedge\; \max(\deg P, \deg Q) = 112 > 100 \\ \wedge\quad & \deg P \nmid \deg Q \;\wedge\; \deg Q \nmid \deg P \\ \wedge\quad & \frac{\partial P}{\partial x}\frac{\partial Q}{\partial y} - \frac{\partial P}{\partial y}\frac{\partial Q}{\partial x} \equiv 1 \\ \wedge\quad & P(0,0) = Q(0,0) = 0 \;\wedge\; \begin{pmatrix} \frac{\partial P}{\partial x}(0,0) & \frac{\partial P}{\partial y}(0,0) \\ \frac{\partial Q}{\partial x}(0,0) & \frac{\partial Q}{\partial y}(0,0) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ \wedge\quad & 2\left(\frac{\partial B}{\partial x}\frac{\partial H}{\partial y} - \frac{\partial B}{\partial y}\frac{\partial H}{\partial x}\right) = 7 H^5 \left(\frac{\partial A}{\partial x}\frac{\partial H}{\partial y} - \frac{\partial A}{\partial y}\frac{\partial H}{\partial x}\right) \quad \text{ahol} \quad H = x^8 y^8 \\ \wedge\quad & \det \begin{pmatrix} \frac{\partial}{\partial x}\left(\frac{P+Q}{\sqrt{2}}\right) & \frac{\partial}{\partial y}\left(\frac{P+Q}{\sqrt{2}}\right) \\ \frac{\partial}{\partial x}\left(\frac{Q-P}{\sqrt{2}}\right) & \frac{\partial}{\partial y}\left(\frac{Q-P}{\sqrt{2}}\right) \end{pmatrix} \equiv 1 \quad \left(\deg \frac{P+Q}{\sqrt{2}} = \deg \frac{Q-P}{\sqrt{2}} = 112\right) \\ \wedge\quad & \mathcal{D}_P := \frac{\partial P}{\partial x}\frac{\partial}{\partial y} - \frac{\partial P}{\partial y}\frac{\partial}{\partial x} \implies \mathcal{D}_P(Q) \equiv 1 \;\wedge\; \forall k \in \mathbb{N}^+ : \deg\big(\mathcal{D}_P^k(x)\big) > \deg\big(\mathcal{D}_P^{k-1}(x)\big) \implies \neg\operatorname{LND}(\mathcal{D}_P) \\ \wedge\quad & \mathcal{B} = \{v_1, v_2\} \subseteq \mathrm{Pl}_\infty : \begin{cases} v_1(x) = -1, v_1(y) = 1 \implies v_1(x^8 y^8) = 0 \\ v_2(x) = 1, v_2(y) = -1 \implies v_2(x^8 y^8) = 0 \end{cases} \implies \#\mathrm{Pl}_\infty(P) = 2 \ne 1 \\ \wedge\quad & S_{P,Q} \supseteq \big\{ (u, v) \in \mathbb{C}^2 \mid v^2 - u^7 = 0 \big\} \ne \emptyset \implies \neg\operatorname{Proper}(F) \\ \wedge\quad & [\mathbb{C}(x,y) : \mathbb{C}(P, Q)] = d \ge 5 \implies \exists z_1 \ne z_2 \in \mathbb{C}^2 : F(z_1) = F(z_2) \implies \neg\operatorname{Inj}(P, Q) \\ \wedge\quad & \neg\operatorname{Inv}(P, Q) \end{aligned} $$ - **solution status counterexample found only**