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Define the polynomial $f(x, y)$ by

$$(x+1)^5+(y+1)^5 = (2x + 6y) ( -x +y )^4 + f(x,y).$$

The curve $C : f(x,y)=0$ has genus 0 and has infinitely many rational points. I don't know if it has infinitely many integral points: suppose not. For integral points on $C$ the identity is:

$$(x+1)^5+(y+1)^5 = (2x + 6y) ( -x +y )^4. \qquad (1)$$

Infinitely many solutions on $C$ with $\gcd(x+1,y+1)=1$ contradict $abc$, so $abc$ implies either finitely many integral solutions, or sufficiently large $\gcd(x+1,y+1)$ (clearing a small gcd will still give abc triples of sufficiently good quality).

The integral points might be growing exponentially, so abc for polynomials doesn't appear to apply.

Q1 Is there a finite extension of $\mathbb{Z}$ where $C$ has infinitely many integral points? How do I find integral points there?

Q2 Is there a similar identity where $f$ is divisible by a quadratic with infinitely many integral points?

I tried to solve Q2 by equating coefficients, but couldn't solve the system.

Other similar identities exist.

If $f$ is divisible by $g= x^2+xy -y^2+1$, integral points on $g$ are consecutive fibonacci numbers $(F_{2n},F_{2n+1})$. If the lhs is still a sum of powers of linear polynomials, it appears unlikely to me that the common factor will always be large, so probably such an identity doesn't exist.

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As to Q1: Your curve has $5$ complex points at infinity, so by Siegel's theorem, for the ring of integers $R$ of any number field there are only finitely many $R$-points.

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  • $\begingroup$ Is this Siegel's theorem that you are using (another source is here)? I am asking because its statement is given only for genus $>0$, whereas the OP claims that his curve has genus $0$. $\endgroup$
    – Alex M.
    Commented Apr 3, 2023 at 14:47
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    $\begingroup$ @AlexM. Unfortunately, both sources you link to miss the genus $0$ case, which itself is interesting and highly nontrivial. If there are sufficiently many rational points on the curve, then this curve has a rational parameterization over the rationals. So the problem essentially reduces to the following: What can we say about a rational function which assumes integer values for infinitely many rational arguments? Siegel proved the necessary condition that this function has at most two poles ... $\endgroup$
    – Peter Mueller
    Commented Apr 3, 2023 at 18:34
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    $\begingroup$ ... Note that indeed this function need not be a polynomial. Take for instance $1/(x^2-2)$. Then its values are integers at $u/v$ where $u,v$ satisfy the Pell equation $u^2-2v^2=1$. These things are handled in Lang's Diophantine Geometry, Chapt. VII, Section 4 (Curves of genus $0$), and probably in his newer edition Fundamentals of diophantine geometry too. $\endgroup$
    – Peter Mueller
    Commented Apr 3, 2023 at 18:38

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