*Question*: Assume that $a,b\in Z$, and $4a^3+27b^2\neq 0$. Prove that there exist infinitely many positive integers $n$ such $n^3+an+b$ is square-free.

I have following

There exist infinitely many postive integers $n$ such $n^2+1$ is squarefree.

**Idea of proof**
Let $S(n)$ be the number of non-squarefree numbers of the form $k^2+1$ less than or equal to $n^2+1$. We can bound it from above by counting every time a number is divisible by the square of a prime:

$S(n) \le \sum_{1 \le p \le n}{\sum_{1 \le k \le n, p^2 \vert k^2+1}{1}}$

Since $k^2+1$ is never divisible by $4$ or $9$, this becomes:

$S(n) \le \sum_{5 \le p \le n}{\sum_{1 \le k \le n, p^2 \vert k^2+1}{1}}$

$-1$ has at most two square roots mod $p^2$, so we can write this as:

$S(n) \le \sum_{5 \le p \le n}{(\frac{2 \cdot n}{p^2}+2)}$

And we can separate the constant term to get:

$S(n) \le 2 \cdot \pi(n) + \sum_{5 \le p \le n}{\frac{2 \cdot n}{p^2}}$

Assuming $n \gt 120$:

$S(n) \le \frac{2 \cdot \phi(120) \cdot n}{120} + 2 \cdot n \cdot \sum_{5 \le p \le n}{\frac{1}{p^2}}$

Simplifying and ignoring that the sum is limited to primes:

$S(n) \le \frac{8 \cdot n}{15} + 2 \cdot n \cdot (\frac{\pi^2}{6} - \frac{1}{16} - \frac{1}{9} - \frac{1}{4} - 1)$

Calculating and rounding up:

$S(n) \le 0.54 \cdot n + 0.45 \cdot n$

$S(n) \le 0.99 \cdot n$

So the density of squarefree numbers of the form $k^2+1$ is at least $0.01$, and therefore there are an infinite number of them.

But I can't prove my Question.