How do I prove that for any polynomial $f(x) \in \mathbb{Z}[x], \operatorname{deg} f \geq 1$, there are infinitely many $n$, such that $f(n)$ is not squarefree?
I have two solutions, one of them is good, correct and quite "straightforward", but the second is very controversial, and I want you to explain to me why it does not work.
$1)$ For large $n, f(n)$ becomes a large integer so is divisible by some prime $p$ say. This means that the polynomial $f(X)$ modulo $p$ factors as $f(X) \equiv(X-n) g(X) \bmod p$. There are two cases to consider.
- $f(n) \equiv 0 \bmod p^{2}$
- $f(n) \not \equiv 0 \bmod p^{2}$ which means that $g(n) \not \equiv 0 \bmod p$.
In the first case $f(n)$ is not square free. In the second case, the derivative $f^{\prime}(x)$ satisfies $f^{\prime}(n) \not \equiv 0 \bmod p$, so we can use Hensel's lemma to find an $n^{\prime} \equiv n \bmod p$ for which
$f\left(n^{\prime}\right) \equiv 0 \bmod p^{2}$. This means that $f\left(n^{\prime}\right)$ is not square free.
I liked this solution, I also have a question: is it necessary to add a restriction on the number $p$ here, it must be taken large enough so that the derivative of the polynomial modulo $p$ does not become identically zero, then everything works?
$2)$ Consider the polynomial $f(x)=c_{k} x^{k}+c_{k-1} x^{k-1}+\ldots+c_{1} x^{1}+c_{0}$, where $f(x) \in \mathbb{Z}[x], \operatorname{deg} f \geq 1$. For any non-squarefree $f(n)$ there are exactly $\operatorname{deg} f$ such $n$ that $f(n)=c_{k} n^{k}+c_{k-1} n^{k-1}+\ldots+c_{1} n^{1}+c_{0}$ – this equality is true (by the fundamental theorem of algebra). This means that since there are infinitely many non-square-free $f(n)$, there also exist infinitely many $n$, that the number $f(n)$ is not square-free.
Best Answer
If $f(x)$ gives only square-free values from some point on, then it should be square-free always. This is because if $d^2\mid f(k)$ for some $k\in \mathbb{N}$ then $d^2\mid f(k+md^2)$. But this cannot happen and here is why.