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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.

vidyarthi
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    @JimmyR. thanks, modified the post – vidyarthi Jan 05 '17 at 08:50
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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 and locally for more information. – Jyrki Lahtonen Jan 05 '17 at 09:19

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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, $$(n^{3/2}+\frac {1}{2}an^{1/2 }-1)^2 <n^3+an^2+bn+c <(n^{3/2}+\frac {1}{2}an^{1/2}+1)^2$$ If $n$ is a perfect square, say $n = m^2$, then the extreme left and right are perfect squares and there is only one perfect square between them, namely $(n^{3/2}+\frac {1}{2}an^{1/2})^2$. Hence if $n=m^2$ and our cubic is a perfect square then we can get $$bm^2+c =\frac {1}{4}a^2m^2$$ For this to hold for sufficiently large $m $, we must have $c=0$ and $b=\frac {1}{4}a^2$. Thus $$n^3+an^2+bn+c =(n^{3/2}+\frac {1}{2}an^{1/2})^2 =(\sqrt {n}(n+\frac {a}{2}))^2$$ which is not a perfect square unless $n $ itself is.