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

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.

• It is impossible for $f(x)$ to both be irreducible, and a product of two or more nontrivial polynomials. – Pace Nielsen Apr 17 '17 at 1:49
• The 4 upvotes must be for perfect formatting, right? – Franz Lemmermeyer Apr 17 '17 at 6:41
• I and others misread the question, apparently automatically converting it to a proper question. As such it of course should be closed. Maybe not deleted, for then Joe Silverman's detailed answer would be gone. – Peter Mueller Apr 17 '17 at 7:47
• @PaceNielsen I've rephrased the question. $y$ was not meant to be a polynomial and I was asking about integer solutions. – trenta3 Apr 17 '17 at 9:53
• The answers given explain why there are only finitely many solutions for any fixed $k>2$. But, the clarified question asks about varying $k$. I don't recall seeing a general result like that, so it may be open. The finiteness follows from the abc conjecture applied to the displayed equation in (the current version of) Joe's answer. – Felipe Voloch Apr 17 '17 at 10:03

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).

• Because of the confusion, it would be useful to say which question you answer – YCor Apr 17 '17 at 9:54

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.

• @W.Schlieper are you confusing integral points and rational points? The number of integral points is finite by Siegel's theorem, even for genus 1. – KConrad Apr 16 '17 at 22:57
• The line $y \in \mathbb Z[x]$ in op's question makes me think he's asking not about points on curves, but about the roots in the polynomial ring. – Anton Fetisov Apr 17 '17 at 1:15
• @AntonFetisov You are right, I misread the question, which as asked of course is nonsense. – Peter Mueller Apr 17 '17 at 7:41
• Thanks for answering what I supposed to be the original question. I've changed the question by only allowing $k \ge 3$ and I'm interested in the whole set of solutions (for any $k$). I know that theorem of Siegel, but it doesn't say anything if you allow $k$ to vary and consider all such solutions. Sorry for mistating the question. – trenta3 Apr 17 '17 at 9:58