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This is hard, so I am looking for partial results and how hard it is.

Let $n>4$. Is it true that the hyperelliptic curve $x^n=y(y+1)$ doesn't have rational point with $x \ne 0$?

If necessarily assume $n$ is prime.

Integral points on the curve are heavily studied.

It is one of the simplest exponential diophantine equations over the rationals.

What tools are used for solving it?

FLT implies there are no solutions with $y$ being $n$-th power.

Added No solution for given $n$ imply Fermat Last Theorem for $n$ as discussed here see (3) with a=1

Assume $u^n-v^n=1$ is counterexample to FLT. Then $(uv)^n=v^n(v^n+1)$ and $(uv,v^n)$ is non trivial rational point on the curve.

joro
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  • Is not this all discussed here? https://mathoverflow.net/q/225324/4312 – Fedor Petrov Aug 29 '18 at 06:40
  • @FedorPetrov I think the curves are the same, but the questions are different. – joro Aug 29 '18 at 08:12
  • Is not it the same question - to prove that there are only trivial rational points? – Fedor Petrov Aug 29 '18 at 08:52
  • @FedorPetrov I am asking about tools and partial results. – joro Aug 29 '18 at 09:27
  • It is essentially equivalent to FLT, so this is equivalent to the tools of studying FLT. – Fedor Petrov Aug 29 '18 at 09:39
  • @FedorPetrov The Fermat curve $x^n+y^n=1$ is not isomorphic to the hyperelliptic curve. For n=5 the computer easily solved the hyperelliptic curve and I don't think there is very simple proof on the Fermat curve. – joro Aug 29 '18 at 11:57
  • What do you mean by "computer easily solved"? Any solution gives a solution of FLT for the same exponent, right? – Fedor Petrov Aug 29 '18 at 13:15
  • The computer showed no solutions. Does the equivalence hold over number fields? I well might be missing something. – joro Aug 29 '18 at 13:40

1 Answers1

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There are no such solutions. Let $x=a/b$ and $y=c/d$ be reduced fractions. Then $a^n/b^n=(c(c+d))/d^2$ and since both sides are reduced fractions we get that $a^n=c(c+d)$ and $b^n=d^2$. From the first equation we deduce that $c=e^n$ and $c+d=f^n$ since $c$ and $c+d$ are co-prime and their product is an $n$-th power. Then $d=f^n-e^n$ so $(f^n-e^n)^2=b^n$. If $n$ is even write $n=2k$ and take square root to get $f^{2k}-e^{2k}=\pm b^k$. If the sign is $+$ then we get $f^{2k}=e^{2k}+b^k$ and since $k>2$ this implies by FLT that one of $e,b,f$ is 0. If $f=0$ then $y=-1$ so $x=0$. $b=0$ is certainly impossible. Finally, $e=0$ means $y=0$. So this option is excluded. If the sign is $-$ we get $e^{2k}=f^{2k}+b^k$ and the same reasoning works.

If $n$ is odd, from $(f^n-e^n)^2=b^n$ we deduce that $b$ is a square so $b=g^2$ and then we can extract the root to get $f^n-e^n= \pm g^n$ and proceed in a way similar to the case where $n$ is even.

S. carmeli
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  • Thanks. Do you think proof without FLT/modularity is possible? I have some partial results: https://mathoverflow.net/questions/332386/elementary-constraints-for-the-solutions-of-zn-2yyz-xn – joro May 24 '19 at 15:30