Let $b(x)=x^4 + 3x^3 + 3x^2 + 2x + 1$, and let $a(x)\in \mathbb Z[x]$ be a separable polynomial. Let $C$ be the plane curve defined by $(y^2+(x+x^2+x^3)a(x))^2-a(x)^2b(x)=0$. I would need to show that if the degree of $a(x)$ is large enough, then $C$ has a finite number of $\mathbb Q$-rational points. My guess is that when the degree of $a(x)$ is large, then $C$ has geometric genus $\geq 2$, so that the claim would follow from Faltings' theorem. However, $C$ might have singularities, so I wouldn't know how to compute, or at least bound from below, its geometric genus. Is there any general principle/result that I can apply? (Notice that unfortunately the curve $y^2=b(x)$ has infinitely many rational points)

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    $\begingroup$ $C$ is a cover of $y^2=b(x)$. What does Riemann-Hurwitz tell you? $\endgroup$ – Felipe Voloch Apr 11 '19 at 19:55
  • $\begingroup$ ...that as soon as the cover is ramified in at least one point, C must have genus at least 2. Thanks a lot for the suggestion! $\endgroup$ – user36371 Apr 12 '19 at 8:42

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