Prove that $\lfloor N/ \lfloor \sqrt{N} \rfloor^2\rfloor = 1$ for $N > 8$

Let $a=\lfloor\sqrt N\rfloor$ then we'll have $a^2\le N< (a+1)^2 = a^2+2a+1$

So we have $N/\lfloor\sqrt N\rfloor^2 \le (a^2+2a)/a^2 = 1+2/a$. So if $a>2$ (ie $a\ge 3$ or $N\ge 9$) we'll have that $N/\lfloor\sqrt N\rfloor^2 < 2$. And of course it will be no less than $1$ so the integral part must be $1$.

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 $N \geq 9$.