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?



Integer polynomial (of degree >1) all of whose values are square-free
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](http://qchu.wordpress.com/2009/09/02/some-remarks-on-the-infinitude-of-primes/)). 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.


