Let $f(x_1,\dots,x_n)$ be squarefree polynomial with integer coefficients. Assume $f$ at integers is not always divisible by a fixed square $m^2 > 1$.
Is it possible $f$ to never be squarefree at integers?
I suspect this is impossible.
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Sign up to join this communityLet $f(x_1,\dots,x_n)$ be squarefree polynomial with integer coefficients. Assume $f$ at integers is not always divisible by a fixed square $m^2 > 1$.
Is it possible $f$ to never be squarefree at integers?
I suspect this is impossible.
This is a supplement to Ilya Bogdanov's answer.
For univariate polynomials Granville (Int. Math. Res. Not. 1998, 991-1009) deduced from the $abc$-conjecture that there are infinitely many natural numbers $x$ such that $f(x)$ is square-free; in fact these numbers have positive density. The same is known unconditionally when $\deg f\leq 3$, by the work of Hooley (Mathematika 14 (1967), 21-26). In addition, Granville's conditional result was extended to multivariate polynomials by Poonen (Duke Math. J. 118 (2003), 353-373).
The answer has been edited, due to a mistake found by joro and explained by Jose Brox --- thanks to both of them! Now we reduce the $n$-variate case to the $(n-1)$-variate one...
The claim that an $n$-variate polynomial will always have a square-free value under our assumptions holds if and only if the same claim holds for an $(n-1)$-variate polynomial (and hence for a univariate polynomial, which still seems to be open). The "only if" part is trivial; let us show the "if" part.
We may assume that $f(\mathbf 0)\neq 0$. For any $p^2\mid f(\mathbf 0)$ there exists $\mathbf x_p$ such that $p^2\nmid f(\mathbf x_p)$. Using CRT, we find an $\mathbf x$ such that $p^2\nmid f(\mathbf x)$ for all $p^2\mid f(\mathbf 0)$. [EDIT] Moreover, there are $n$ sets of the form $S_i=P\mathbb Z+a_i$ such that every $\mathbf x\in\prod_{i=1}^n S_i$ works.
Consider now the polynomial $g(x_1,x_2,x_3,\dots,x_{n-1},t)=f(x_1,\dots,x_{n-1},tx_{n-1})$. It is also square-free (when passing from $g$ to $f$ by substituting $t=x_n/x_{n-1}$, a square may disappear only if it was $x_{n-1}^2$; this cannot happen since $g(\mathbf 0)=f(\mathbf 0)\neq0$). Next, there are only finitely many values $\alpha$ such that $g(x_1,\dots,x_{n-1},\alpha)$ is not square-free: at these exceptional $\alpha$ the discriminant of $g$ with respect to one of the $x_i$ should vanish, and all these discriminants are not identically zero.
Now we choose $\alpha\in S_nS_{n-1}^{-1}$ for which $f_1(x_1,\dots,x_{n-1},\alpha)$ is square-free; let $\alpha=k_n/k_{n-1}$ with $k_i\in S_i$. Then the polynomial (with integer coefficients) $h(x_1,\dots,x_{n-2},x_{n-1})=f(x_1,\dots,x_{n-2},k_{n-1}x_{n-1},k_nx_{n-1})=g(x_1,\dots,k_{n-1}x_{n-1},\alpha)$ is square-free, and for $k_i\in S_i$ the values $h(k_1,\dots,k_{n-2},1)=f(k_1,\dots,k_n)$ and $h(\mathbf 0)=f(\mathbf 0)$ have no common square divisors (other than 1). By the $(n-1)$-variate case, $h$ has a square-free value, therefore so does $f$.