<s>[INST] Prove that $\lfloor N/ \lfloor \sqrt{N} \rfloor^2\rfloor = 1$ for $N > 8$ 
Prove that $\lfloor N/ \lfloor \sqrt{N} \rfloor^2\rfloor = 1$ for $N > 8$.

This is simple to check by listing and the only exceptions are for $N = 2,3,8$.  I am looking for a simple proof if possible.  This problem arises from solving polynomials with constraints on the range of coefficients.
 [/INST] By definition of floor,
$$
\sqrt{N}-1\lt\left\lfloor\sqrt{N}\right\rfloor\le\sqrt{N}\tag1
$$
Therefore,
$$
\begin{align}
1\le\frac{N}{\left\lfloor\sqrt{N}\right\rfloor^2}&\lt\frac{N}{\left(\sqrt{N}-1\right)^2}\\
&=2-\frac{\left(\sqrt{N}-2\right)^2-2}{\left(\sqrt{N}-1\right)^2}\\[6pt]
&\lt2\tag2
\end{align}
$$
when $N\gt\left(2+\sqrt2\right)^2=6+4\sqrt2\doteq11.656854$, so we just need to verify $(2)$ for $N\in\{9,10,11\}$ by hand. Because $\frac99,\frac{10}9,\frac{11}9\lt2$, we have shown that for $N\gt8$,
$$
1\le\frac{N}{\left\lfloor\sqrt{N}\right\rfloor^2}\lt2\tag3
$$
</s>