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In my previous post I asked about the hyperbolicity of the affine surface $S'=\{zw \neq o\}$ in the projective surface $z^2 = P(x) Q(y)$ in $\mathbb{P}^3$, where $P$ and $Q$ are two general polynomials of the same degree $d$. Jason wrote that this affine surface is a finite unbranched cover of the product of two affine hyperbolic curve $(\mathbb{C}\setminus Z(P(x)))\times (\mathbb{C}\setminus Z(Q(y)))$,$(x,y,z) \mapsto (x/z, y/z)$ and therefore $S'$ is also hyperbolic as well (what is the generic condition on polynomials $P$ and $Q$?).//

additionally I could prove that this surface is a surface of general type if $d\ge 5$ and the roots of $P$ and $Q$ are not the same.//

Now assume the equation $z^2 = P(x) Q(y)$ on a number field $K$. I want to ask if the number of $K$ - rational points in the affine part $zw \neq 0$ is finite? I want to deduce this result from Lang conjectures which asserts that:

on a surface of general type the Zariski closure of rational points is a proper sub variety not of general type, i.e. in this case it is a finite union of rational and elliptic curves and a finite singular points.

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  • $\begingroup$ The map is $(x,y,z)\mapsto (x,y)$, not $(x,y,z)\mapsto (x/z,y/z)$. $\endgroup$ Apr 1, 2014 at 12:12
  • $\begingroup$ How much is your degree $d$? If too small, the variety is certainly not of general type. Why do you think that you dont have rational curves on the surface? $\endgroup$ Apr 1, 2014 at 12:28
  • $\begingroup$ thanks Jason, What about the generic condition on the polynomials $P,Q$?Is it always true for all polynomials $P,Q$. $\endgroup$
    – Mehdi
    Apr 1, 2014 at 12:29
  • $\begingroup$ If you take $P$ and $Q$ of small degree (for example of degree $1$), your surface $z^2=P(x)Q(y)$ is of course rational. That's why I asked what you assume on $P$, $Q$ to say that your variety is of general type. $\endgroup$ Apr 1, 2014 at 12:46
  • $\begingroup$ $d$ can be chosen arbitrary large. $\endgroup$
    – Mehdi
    Apr 1, 2014 at 12:52

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It depends on the polynomials $ P $ and $ Q $ if they admit a polynomial parametrization they can even have infinitely many integer solutions.

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  • $\begingroup$ My question is for general polynomials $P$ and $Q$, i.e. the coefficients of $P, Q$ satisfy in some generic condition. $\endgroup$
    – Mehdi
    Apr 1, 2014 at 16:13
  • $\begingroup$ Note that in my previous post it was shown that the affine part $z\neq 0$ is hyperbolic so it is natural (regarding Lang Conjecture) to expect that there is finite rational points on this part. $\endgroup$
    – Mehdi
    Apr 1, 2014 at 16:18

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