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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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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).

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    $\begingroup$ You can find the paper of Granville here. $\endgroup$ – Jeremy Rouse Sep 23 '17 at 16:02
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    $\begingroup$ So it would be accurate to say that we do not know definitively for degree greater than $3$ but do very strongly suspect that it is indeed impossible. $\endgroup$ – Aaron Meyerowitz Sep 23 '17 at 17:18
  • $\begingroup$ @AaronMeyerowitz: Yes. $\endgroup$ – GH from MO Sep 23 '17 at 17:54
  • $\begingroup$ There is alleged counterexample to Ilya's answer, check the comments. $\endgroup$ – joro Sep 25 '17 at 6:15
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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 $f$.

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  • $\begingroup$ See also my response. $\endgroup$ – GH from MO Sep 23 '17 at 16:01
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    $\begingroup$ Isn't this counterexample: $f(x,y,z)=(x^2+2)^2+y-z$. The bad prime is $2$ and $f(1,1,1)=3^2$. Then $f(t,t,t)=(t^2+2)^2$ which is never squarefree. $\endgroup$ – joro Sep 24 '17 at 7:45
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    $\begingroup$ As @joro shows, you need also to assure that $g(t)$ is squarefree as a polynomial. $\endgroup$ – Jose Brox Sep 24 '17 at 14:12
  • $\begingroup$ @JoseBrox I am not sure yours is enough, g(t) may always be divisible by square even if it and its content are squarefree like $(x-1)(x-2)(x-3)(x-4)$ $\endgroup$ – joro Sep 24 '17 at 15:45
  • $\begingroup$ @joro What Ilya Bogdanov was saying is that the multivariate case can be reduced to the univariate case, since if $f(x_1,\ldots,x_n)$ is square-free and without a divisor $m^2>1$ then, he claims, we can find a polynomial in one variable $g(t)$ whose image is inside the image of $f$, so that if we know the result to be true for one variable, then it would be true also for several variables. But his reasoning above has a flaw, as you have shown: from $f$ he has got a polynomial $g(t)$ which satisfies it has no $m^2>1$ when $f$ does, but he has forgotten to guarantee that $g(t)$ is squarefree! $\endgroup$ – Jose Brox Sep 24 '17 at 16:11

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