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Given any positive integer $n>1$, it is fairly straightforward to prove that there exists at least one (and perhaps many) square integers lying strictly between $4n^3$ and $4n^3 + 4n^2$. (Proof: This is equivalent to the claim that there is an integer between $2n\sqrt{n}$ and $2n\sqrt{n+1}$, which is true because $2n\sqrt{n+1} - 2n\sqrt{n} > 1$.) Does anybody see a way to find an explicit formula (in terms of $n$, preferably as the square of a polynomial) for such a square integer?

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    $\begingroup$ @NickGuerrero correct, I neglected to include the condition $n>1$. That's fixed now. $\endgroup$ – mweiss Nov 3 '16 at 5:25
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    $\begingroup$ @Masacroso The original post omitted the coefficient on $n^2$, which I corrected. But yes, my question is precisely how to solve that inequality. $\endgroup$ – mweiss Nov 3 '16 at 5:26
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    $\begingroup$ Ah, that makes more sense then. How about $\lfloor{\sqrt{4n^3+4n^2}}\rfloor$? $\endgroup$ – Nick Guerrero Nov 3 '16 at 5:30
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    $\begingroup$ I don't think you can get such a polynomial. As a function of $n$, this square grows like $n^3$. The square of a polynomial in $n$ grows like $n$ to an even power. $\endgroup$ – Matthew Conroy Nov 3 '16 at 5:37
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    $\begingroup$ @MatthewConroy I think you have distilled the essence of the question and explained why what I am looking for doesn't exist. If you would like to post it as an answer, I will accept it. $\endgroup$ – mweiss Nov 3 '16 at 13:19
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Let's suppose we have a polynomial $r$ such that $r(n)^2$ is a perfect square with $$4n^3 \le r(n)^2 \le 4n^3+4n^2$$ for all $n$. Suppose $r(x)$ has degree $k$. Then $$ \frac{4n^3}{n^{2k}} \le \frac{r(n)^2}{n^{2k}} \le \frac{4n^3}{n^{2k}}+\frac{4n^2}{n^{2k}}.$$ Letting $n$ go to infinity, the middle expression approaches a non-zero constant, and we conclude that $2k\ge3$ and $2k\le3$, i.e., $k=\frac{3}{2}$. Since $k$ is an integer, this is a contradiction. Hence, no such polynomial $r$ exists.

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  • $\begingroup$ Moreover the same argument (with minor modifications) shows that $r(n)$ can't be a rational function, either. $\endgroup$ – mweiss Nov 3 '16 at 23:39

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