$\lfloor \sqrt{\lfloor x\rfloor}\rfloor = \lfloor \sqrt{ x}\rfloor$
$\lceil \sqrt{\lceil x\rceil}\rceil = \lceil \sqrt{x}\rceil$
continuous, monotonically increasing $f$ with property $f(x)\in Z \Rightarrow x \in Z$. Then below holds $$ \begin{align} \lfloor f(\lfloor x\rfloor)\rfloor = \lfloor f(x)\rfloor,\ \lceil f(\lceil x\rceil)\rceil = \lceil f(x)\rceil \end{align} $$
$\lceil \sqrt{\lfloor x\rfloor}\rceil = \lceil \sqrt{ x}\rceil$, only when $m^2 < x < m^2+1$ not hold.
$$ \begin{align} \sum_{k=0}^{m-1} \lfloor \frac{n+k}{m} \rfloor =n= \sum_{k=0}^{m-1} \lceil \frac{n-k}{m} \rceil \end{align} $$$$ \begin{align} \sum_{k=0}^{n-1} \lfloor \sqrt{k} \rfloor = na - \frac{1}{3} a^3 - \frac{1}{2}a^2 - \frac{1}{6}a,\ a = \lfloor \sqrt{n} \rfloor \end{align} $$$$ \begin{align} \sum_{k=0}^{m-1} \lfloor \frac{nk+x}{m} \rfloor = d \lfloor \frac{x}{d} \rfloor + \frac{(m-1)(n-1)}{2} + \frac{d-1}{2} = \sum_{k=0}^{n-1} \lfloor \frac{mk+x}{n} \rfloor,\ d = \text{gcd}(m,n) \end{align} $$
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