<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] Suppose $N = n^2 + k$ such that $0 \leq k < 2n+1$ so that we have $n^2 \leq N < (n+1)^2$. Then $\lfloor \sqrt{N} \rfloor = n$.
Now, we can simplify as below:
$$
\Bigg\lfloor \frac{N}{{\lfloor \sqrt{N} \rfloor}^2} \Bigg\rfloor
= \bigg\lfloor \frac{n^2 + k}{n^2} \bigg\rfloor
= 1 + \bigg\lfloor \frac{k}{n^2} \bigg\rfloor
$$
Since $k < 2n + 1$, we also have
$$ \frac{k}{n^2} < \frac{2n + 1}{n^2} = \frac{2}{n} + \frac{1}{n^2} $$
Thus, $1 + \Big\lfloor \frac{k}{n^2} \Big\rfloor = 1 + \Big\lfloor \frac{2}{n} \Big\rfloor$ and for this to be $1$, we need $\Big\lfloor \frac{2}{n} \Big\rfloor = 0$ i.e. $n \geq 3$ or equivalently $N \geq 9$.
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