<s>[INST] Do there exist infinitely many $n$ such that $n^3+an+b$ is squarefree? 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.
 [/INST] Erdos was the first to show that cubic polynomials (meeting obvious necessary conditions) do take on square-free values.  More generally he considered polynomials of degree $\ell$ taking on $\ell-1$-th power-free values infinitely often.  Granville has shown that the ABC conjecture allows one to find square-free values of higher degree polynomials. See also interesting recent work by Reuss who considers $\ell-1$-th power-free values of degree $\ell$ polynomials evaluated at primes, and also gives many other references. 
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