Square free part of $(n^{2} \pm 1)$ misses some square-free number? For convenience, let rad$(n)$ denote the square-free part of a positive integer (This is unusual notation, since rad$(n)$ is also used to denote product of primes dividing $n$). I was wondering if there is a square-free integer which is not equal to either rad$(n^{2}-1)$ nor rad$(n^{2}+1)$, regardless of input value $n$?
The reason I chose $(n^{2}\pm 1)$ is because for any polynomial $f(n)$ of degree $\geq 3$, the rad$(f(n))$ perhaps grows too fast and hence misses a lot of square-free integers. Albeit, I have not formally proved this. The linear case is not much interesting. I though the easiest degree $2$  polynomials would be of the form $(n^{2} \pm 1)$.
Therefore, I decided that may be taking the square-free part of the pair of polynomials $(n^{2}+1)$ and $(n^{2}-1)$ would become interesting. Any hints/thoughts would be much appreciated, since I have never read about such problems.

Updated after your edit clarifying what you mean by $\mathrm{rad}(n)$:  no; in fact, every squarefree integer $m > 1$ appears as $\mathrm{rad}(n^2 - 1)$.  Proof:  given $m > 1$ squarefree, the Pell equation $x^2 - my^2 = 1$ has infinitely many integer solutions $(x, y)$.  Given such a solution, we have $x^2 - 1 = my^2$, so $\mathrm{rad}(x^2 - 1) = m$.  (The theorem on Pell's equation fails if $m = 1$, but of course we have $1 = \mathrm{rad}(0^2 + 1)$.)
Original answer:  Let $p$ be a prime that is not of the form $n^2 \pm 1$ for any $n$; examples are $7, 11, 13, 19$, and so on.  By Mihăilescu's theorem (a big hammer, I know), we cannot have $p^k = n^2 \pm 1$ for any $k > 1$, so we cannot have $p = \mathrm{rad}(n^2 \pm 1)$.
