# Finite set of (perfect power) polynomial values?

First of all excuse my ignorance in number theory, the following question might have a well-known solution or it might be an open problem, I just don't know enough in that area of mathematics (and many others). Let $P\in \mathbb{Z}[X]$ irreducible and of degree at least 1. For $k\in \mathbb{N}, k\geq 2$, denote by $S_k$ the set of integers $n$ such that there exists $m\in \mathbb{Z}$ and $P(n)=m^k$. Is $S_k$ finite?

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No. For example, by the theory of Pell equations many quadratic polynomials such as 2x^2 + 1 take on square values infinitely often. – Qiaochu Yuan Mar 23 '10 at 23:28
On the other hand, if either k or the degree of P is greater than or equal to 3, see en.wikipedia.org/wiki/Siegel%27s_theorem_on_integral_points . – Qiaochu Yuan Mar 24 '10 at 0:16
Thank you Qiaochu, I can close this topic then. – Portland Mar 24 '10 at 0:41

As Qiaochu said in the comments, you must include Pell type equations as a special case, because they are the only counter example. At least for $k=2$, Siegel's theorem on integral points on algebraic curves implies that if your polynomial $P(x)$ has at least three distinct roots then $P(n)=m^2$ has only finitely many solutions. So your conjecture is, in particular, true for irreducible polynomials of degree higher than 2.