On the solutions of $f(x) = y^k$ with $f \in \mathbb{Z}[x]$, $k \in \mathbb{N}$

Question

I was wondering if it is true that the set of integer solutions of the equation $$ f(x) = y^k $$ is finite, where $f$ is an irreducible integer polynomial of degree $d \ge 2$ and $y \in \mathbb{Z}$, $k \in \mathbb{N}$ with $k \ge 3$.

That is, given an integer irreducible polynomial $f$ of degree greater than two, if the set $\{ (x, y, k) \in \mathbb{Z}^3 | k \ge 3, f(x) = y^k \}$ is finite.

I searched in the literature but could not find the answer, and trying to prove it myself has proved unsuccesful.

The MO question Polynomials which always assume perfect power values asks the question for the same equation but assumes that it holds for every integer number, so it isn't quite the solution to current question.

Thanks in advance for any help or counterexample.

Ok, due to me not paying lot of attention to how i've written it down, the question has been misread a lot, so I've rephrased the question. I hope it is more clear now.

Answer 1

To expand a bit on Peter Mueller's answer, Siegel's theorem isn't overkill, although this case is a bit easier. When $k=2$, these curves are classically called hyperelliptic curves, and for higher $k$ they were named superelliptic curves by Serge Lang. Here's a rough sketch of the proof of finiteness for $k\ge3$, but you can find a full proof in many books. Factor $f(x)$, so $y^k=c\prod(x-\alpha_i)$. Then in any solution, the quantities $x-\alpha_i$ are more-or-less relatively prime, so they are more-or-less $k$'th powers. More precisely, letting $K$ be the splitting field of $f(x)$ and $R_K$ its ring of integers, there is a finite set of ideals $\{\mathfrak a_1,\ldots,\mathfrak a_t\}$ so that every $(x-\alpha_i)R_K$ is one of the $\mathfrak a_i$ times the $k$'th power of an ideal. Using finiteness of class number, one finally get $x-\alpha_i=b_iz_i^k$, with $b_i$ chosen from a finite set and $z_i\in R_K$. In particular, we get $$\alpha_2 - \alpha_1 = b_2z_2^k - b_1z_1^k.$$ It follows (after further work) that $z_2/z_1$ is a very good approximation to $(b_1/b_1)^{1/k}$, and now an application of Roth's theorem (or one of the weaker variants, the one due to Thue will suffice) gives finiteness of solutions. However, this gives an ineffective bound for the largest solution, although it does give an effective bound for the number of solutions. You can use Baker's theorem on linear forms in logarithms to get an effective bound for the largest solution.

As you can see from this sketch, this is a hard theorem, since it relies on a bunch of fairly deep results (finiteness of class number, finite generation of the unit group, and Diophantine approximation to algebraic numbers).

Answer 2

For $d=k=2$ this isn't true. For instance take $f(x)=2x^2+1$, the Pellian equation $y^2-2x^2=1$ has infinitely many integral solutions.

If $d\ge3$ or $k\ge3$, then the curve given by $f(X)-Y^k=0$ has positive genus, and such curves have (by an old theorem of Siegel, a precursor of Falting's theorem) only finitely many integral points.

Maybe Siegel's theorem is overkill here. On the other hand, even the elliptic curve case $d=3$, $k=2$ isn't easy.