*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.**
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> 
> 


**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.



Do there exist infinitely many $n$ such that $n^3+an+b$ is squarefree?
[Erdos](https://www.renyi.hu/~p_erdos/1953-02.pdf) 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](http://www.dms.umontreal.ca/~andrew/PDF/polysq3.pdf) has shown that the ABC conjecture allows one to find square-free values of higher degree polynomials. See also interesting recent work by [Reuss](http://arxiv.org/pdf/1307.2802v1.pdf) who considers $\ell-1$-th power-free values of degree $\ell$ polynomials evaluated at primes, and also gives many other references. 


