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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
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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
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up vote 4 down vote accepted

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.

Anyway, for the general question for any exponent it's better that you read the full story in "The diophantine equation f(x)=g(y)" by Y. Bilu and R.F. Tichy.

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Oops! I should have updated the comments section... –  Gjergji Zaimi Mar 24 '10 at 0:42
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