Is there an integer polynomial $ A \in {\mathbb Z} [ X ]$ of degree $d\geq 2$ such that for any integer $n\in {\mathbb Z}$ , $ A(n) $ is a square-free integer?
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No. WLOG $A$ is irreducible. Pick a sufficiently large prime $p$ dividing $A(k)$ for some $k$ (there are infinitely many such primes, for example by the argument here). In particular pick $p$ large enough so that it is relatively prime to the coefficients of $A$ and to $d$, and so that it does not divide the discriminant of $A$. Then the congruence $A(x) \equiv 0 \bmod p^2$ has a solution by Hensel's lemma.
answered Nov 6, 2010 at 10:11
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5$\begingroup$ If there are only n primes dividing A(k) as k varies, there are only 2^n possibilities for squarefree values of A(k), and then you can use the fact that the polynomial is unbounded. $\endgroup$ Commented Nov 6, 2010 at 11:14
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$\begingroup$ Thanks Qiaochu. If I had been more familiar with p-adic analysis, I probably wouldn't have needed to ask the question here. $\endgroup$ Commented Nov 6, 2010 at 12:30