Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider the identity

$$(x+1)^5+(y+1)^5 = (2x + 6y) ( -x +y )^4 + f(x,y)$$ where $f(x,y)=-x^5 + 2*x^4*y + 12*x^3*y^2 - 28*x^2*y^3 + 22*x*y^4 - 5*y^5 + 5*x^4 + 5*y^4 + 10*x^3 + 10*y^3 + 10*x^2 + 10*y^2 + 5*x + 5*y + 2$.

The curve $C : f(x,y)=0$ is 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.

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.