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.
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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 $1$.
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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 $N\gt8$, $$ 1\le\frac{N}{\left\lfloor\sqrt{N}\right\rfloor^2}\lt2\tag3 $$
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 $N \geq 9$.