Problem 1.
Find all real number a  such that for any positive integer n,
⌊α⌋+⌊2α⌋+⋯+⌊nα⌋
is a multiple of n
(Note that ⌊z⌋ denotes the greatest integer less than or equal to z.For example, ⌊-π⌋=-4 and  ⌊2⌋=⌊2.9⌋=2.)

 

Solution
Let $f_{k}(\alpha) =\lfloor{k\alpha}\rfloor$ and $N = \sum_{k=1}^n f_k(\alpha)$. If $\alpha$ is a solution to the problem, so is $\alpha + 2m$, where $m$ is an arbitrary integer. This is because $\lfloor{k(\alpha + 2m)}\rfloor = 2km + \lfloor{k\alpha}\rfloor$ and if $n | N$ then
$n | \sum_{k=1}^n f_k(\alpha + 2m) = 2km + \sum_{k=1}^n \lfloor{k\alpha}\rfloor = N + \frac{2mn(n+1)}{2} = N + mn(n+1)$.

This means we can limit our findings on an arbitrary interval of length 2 and extrapolate from there.

For sake of simplicity take the $\alpha \in [-1,1)$. We split the problem into two cases:

Case 1: Interval [-1,0)
This means $-1\leq \alpha < 0 \implies -n \leq n\alpha < 0 \implies \lfloor{n\alpha}\rfloor \in \{-n,-n+1,\dots,-2,-1\}$

Claim: $\lfloor{n\alpha}\rfloor = -1$ for all positive integers n.

Using strong induction, base case is trivial $1 | \lfloor{\alpha}\rfloor = -1$. Assume $\lfloor{n\alpha}\rfloor = -1$ for all $n \in \{1,2,\dots,k-1,k \}$

Consider $n = k+1$: $n = k+1 | -k + \lfloor{(k+1)\alpha}\rfloor$.
Now $ \lfloor{(k+1)\alpha}\rfloor \in \{-(k+1),-k,\dots,-2,-1\}$ The only way for $k+1$ to divide $-k + \lfloor{(k+1)\alpha}\rfloor$ is for $\lfloor{(k+1)\alpha}\rfloor = -1$. This concludes the proof.


Now, since $\lfloor{n\alpha}\rfloor = -1$, $-1\leq n\alpha < 0$. Thus $-\frac{1}{n}\leq \alpha < 0$. As the problem's condition holds true for any positive integer n, it must hold for arbitrarily large n, i.e. on the limiting case as n tends to infinity. Taking the limit on our inequality yields $0\leq \alpha < 0$, with no $\alpha$ satisfying the double inequality.

Case 2: Interval [0,1)
This means $0\leq \alpha < 1 \implies 0 \leq n\alpha < n \implies \lfloor{n\alpha}\rfloor \in \{0,1,\dots,n-2,n-1\}$.

Claim: $\lfloor{n\alpha}\rfloor = 0$ for all positive integers n.

Using strong induction, base case is again trivial $1 | \lfloor{\alpha}\rfloor = 0$. Assume $\lfloor{n\alpha}\rfloor = 0$ for all $n \in \{1,2,\dots,k-1\}$

Consider $n = k+1$: $n = k+1 | 0 + \lfloor{(k+1)\alpha}\rfloor$.
Now $\lfloor{(k+1)\alpha}\rfloor \in \{0,1,2,\dots,k-1,k\}$ The only way for $k+1$ to divide $\lfloor{(k+1)\alpha}\rfloor$ is for $\lfloor{(k+1)\alpha}\rfloor = 0$. This concludes the proof.

Since $\lfloor{n\alpha}\rfloor = 0$ for any n, it implies $\alpha = 0$.

Thus the only reals $\alpha$ satisfying the problem are $\alpha = 2m$, where m is an integer.