In a question I made few years ago, asking if $\frac{a^n+b^n}{a+b} = c^n$ can have a solution for $a,b,c,n\in \mathbb N$, $a^n+b^n\neq a+b$, and $\gcd(a+b,c)=\gcd(a,b)=1$, the user @MummytheTurkey conjectured that there were no non-trivial solutions for $n>3$, mentioning an heuristic of why the conjecture might be true (citation follows):

I only thought about the case when $n$ is odd. In that case the LHS is a polynomial of degree $n−1$. This equation then cuts out a curve in $\mathbb{P}(n,n,n−1)$ (weighted projective space). Solutions should conincide(ish) with rational points on this curve (call it $C_n$). When $n\geq 5$ this curve has genus $\geq 2$, so has only finitely many rational points by Faltings' Theorem. In practice when one searches for points on curves of genus $\geq 2$ with nice enough equations you usually find all the solutions straight away (at least if you use clever enough point searching)

I would like to know which strategy (if any) could be followed to show that this conjecture is indeed true (or false, for instance a strategy to look for a counterexample).

Thanks for your time!

  • $\begingroup$ What I said about this defining a curve is nonsense, the weighting is not reasonable enough. I'll leave a few comments below which I think could salvage the heuristic, but beware I am far from an expert on this topic. An idea could be to follow a strategy which I adapt from a paper of Poonen--Schaefer--Stoll [PSS] ($x^2 + y^3 = z^7$). $\endgroup$ Apr 5 at 14:12
  • $\begingroup$ (cont) Consider the scheme $S \subset \mathbb{A}^3$ given by the vanishing of $f = a^n + b^n - c^n(a + b)$. Clearly $S$ admits an action of $\mathbb{G}_m$ by $(a,b,c) \mapsto (u^n a, u^n, b, u^{n-1} c)$. One should then consider the quotient stack $\mathscr{S} = [S/\mathbb{G}_m]$. There is a stack-theoretic Euler characteristic $\chi(\mathscr{S})$ (which mirrors $\chi = 2-2g$ from the scheme case) which one can hope to compute. The (heuristic, maybe theorem??) being that $\chi(\mathscr{S}) < 0$ $\endgroup$ Apr 5 at 14:13
  • $\begingroup$ (cont) implies finitely many integer points, and $\chi(\mathscr{S}) > 0$ implies infinitely many integer points (if there is one). Note that if the action were weighted $(n,n,n)$ this is Faltings' theorem plus the fact that conics have infinitely many rational points if they have one. The basic strategy should be to (1) compute the algebra of invariants $(\mathbb{Z}[a,b,c]/(f))^{\mathbb{G}_m}$ by the above weighted action of $\mathbb{G}_m$ on $S$ in order to determine the "coarse space" for $\mathscr{S}$. One then computes the Euler characteristic of $\endgroup$ Apr 5 at 14:13
  • $\begingroup$ (cont). $\mathscr{S}$ using the data of the genus of the coarse space and ramification data (see [PSS, 3.2]). One then has a Heuristic for finiteness of integer points, maybe there is a descent argument to compute them -- (hard?). $\endgroup$ Apr 5 at 14:13
  • $\begingroup$ (cont). Finally, one can search for counterexamples by just pointsearching on $S$ using efficient algorithms such as Magma's PointSearch. Of course, none of this bars that there is a much easier way to do things..... $\endgroup$ Apr 5 at 14:31


You must log in to answer this question.

Browse other questions tagged .