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Can we prove that for any $a, b ,c$, there exists an integer $n$ such that $\sqrt{n^3+an^2+bn+c}$ is not an integer?

I think yes, because the range of polynomial is the whole of $\mathbb{R}$. Any ideas. By the way, is this related to elliptic curves by any chance? Thanks beforehand.

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    @JimmyR. thanks, modified the post – 2017-01-05
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    If $a,b,c$ are integers such that $x^3+ax^2+bx+c$ has only simple zeros, then this is related to integer points of the elliptic curve $y^2=x^3+ax^2+bx+c$. A theorem due to Siegel says that there are only finitely many points with integer coordinates. See [Wikipedia](https://en.wikipedia.org/wiki/Elliptic_curve#Integral_points) and [locally](http://math.stackexchange.com/q/32847/11619) for more information. – 2017-01-05

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This is a $1998$ putnam B6 problem.

An alternate approach: We write the assumed perfect square $n^3+an^2+bn+c $ in the form $(n^{3/2 }+ dn^{1/2}+f)^2$ giving us $$n^3+an^2+bn+c =n^3+2n^2d +2 (\sqrt {n})^3f+nd^2+2d\sqrt {n}f +f^2$$ Choosing $d=\frac {1}{2}a$ and $f=\pm 1$, we then, for $n $ sufficiently large, $$(n^{3/2}+\frac {1}{2}an^{1/2 }-1)^2

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    how do we know that we can express our assumed square as $(n^{3/2}+dn^{1/2}+f)^2$? – 2017-01-05