ceiling-and-floor-functionsdivisibilitynumber theory
I tried to look at the cases when $n$ is a perfect square. Then $\left\lfloor \sqrt{n} \right\rfloor^{2} +2= n+2$, $ n^2 + 1 =(n-2)(n+2) + 5$. Then we must have $(n+2)|5$.
But only $1$ and $5$ divide $5$. Thus, $n=3$, but that is not a solution since we assumed $n$ to be a perfect square. The problem therefore has no perfect-square solutions.
I'm not sure how relevant this is to the general case, but I did not manage to get any further.
Thank you for your help in advance.
Your original statement is not true (please read the whole answer). But the following statement is correct:
$$\left \lfloor {\log_2 n} \right \rfloor = \left \lfloor {\log_2 \left \lceil \frac {n-1}{2}\right \rceil} \right \rfloor + 1\tag{1}$$
Proof:
Every number $n$ can be placed between two cosecutive powers of 2. In other words, there exists $k$ such that:
$$2^k\le n\le2^{k+1}-1\tag{1}$$
Obviously:
$$k\le\log_2n<k+1$$
$$k=\lfloor\log_2n\rfloor\tag{2}$$
On the other side from (1):
$$\frac{2^k-1}2 \le \frac{n-1}{2} \le \frac{2^{k+1}-2}2$$
$$2^{k-1}-\frac12 \le \frac{n-1}{2} \le 2^k-1$$
$$\lceil 2^{k-1}-\frac12\rceil \le \lceil\frac{n-1}{2}\rceil \le 2^k-1$$
$$2^{k-1} \le \lceil\frac{n-1}{2}\rceil \le 2^k-1$$
$${k-1} \le \log_2\lceil\frac{n-1}{2}\rceil \le \log_2 (2^k-1)<k$$
$${k-1} =\lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor$$
$$k=\lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor+1\tag{3}$$
By comparing (2) and (3) you get:
$$\lfloor\log_2n\rfloor = \lfloor \log_2\lceil\frac{n-1}{2}\rceil \rfloor+1$$
...which completes the proof.
You can easily prove that the original statement is not true. You are basically saying that the function:
$$f(n)=\left \lfloor {\log n} \right \rfloor - \left \lfloor {\log \left \lceil \frac {n-1}{2}\right \rceil} \right \rfloor - 1$$
...is equal to zero for all values of $n$.
This is not true if "$\log$" stands for logartihm with base 10:
This is also not true if "$\log$" stans for natural logarithm "$\ln$":
If you don't trust these plots, calculate the value for $n=45$ and in both cases the result is -1, not 0.
Let $\lfloor n(a - \lfloor \sqrt a \rfloor) \rfloor = r$ and $\lfloor a \rfloor + 1 = k^2$, we have that $$k^2 + \frac{r}{n} \le a + 1 < k^2 + \frac{r + 1}{n} \implies \sqrt{a + \frac{n - r - 1}{n}} < k \le \sqrt{a + \frac{n - r}{n}}$$
Deducing that $$\left\lfloor\sqrt{a + \frac{i}{n}}\right\rfloor = \left\{ \begin{align} k - 1 &\text{ where } j < n - r\\ k &\text{ where } n - r \le j \end{align} \right.$$
This implies that $$\sum_{i = 0}^{n - 1}\left\lfloor\sqrt{a + \frac{i}{n}}\right\rfloor = (n - r)(k - 1) + rk = n(k - 1) + r$$
Since $k - 1 = \lfloor\sqrt {a} \rfloor$, $$\sum_{i = 0}^{n - 1}\left\lfloor\sqrt{a + \frac{i}{n}}\right\rfloor = n\lfloor a \rfloor + \lfloor n(a - \lfloor \sqrt a \rfloor) \rfloor$$
Best Answer
Let $m^2=n-k$ be the largest square less than or equal to $n$.
$\implies0\le k\le 2m$
We have $n-k+2|n^2+1$ $$\implies n-k+2|1+n(k-2)$$ $$\implies n-k+2|k^2-4k+5$$ $$\implies m^2+2|k^2-4k+5$$ $\dfrac{k^2-4k+5}{m^2+2}=1,2\text{ or }3$ otherwise the inequality $0\le k\le 2m$ is violated.
If $\dfrac{k^2-4k+5}{m^2+2}=1$ we then have $(k-m-2)(k+m-2)=1$.
The only solutions for the above equality are $(m,k)=(0,1),(0,3)$, but $k>2m$, so no solution exists.
$\dfrac{k^2-4k+5}{m^2+2}=2$ we then have $(k-2)^2+1=2m^2+4$. $k$ should be odd. The LHS is of the form $8k+2$ while RHS is of the form $8k+4$ or $8k+6$. So no solution exists.
If $\dfrac{k^2-4k+5}{m^2+2}=3$ we then have $(k-2)^2+1=3m^2+6$. $3$ never divides the LHS so there are no solutions.