This was also P4 from the 2022 Azerbaijan BMO TST and it was proposed by Serbia. Try to find an idendity in structure of the question[/hide], Also in my opinion this question was really hard and no one in TST could fully solve this

BMO Shortlist 2021/N5 wrote: A natural number $n$ is given. Determine all $(n - 1)$-tuples of nonnegative integers $a_1, a_2, ..., a_{n - 1}$ such that \[ \left \lfloor \frac{m}{2^n - 1} \right \rfloor + \left \lfloor \frac{2m + a_1}{2^n - 1} \right \rfloor + \left \lfloor \frac{2^2 m + a_2}{2^n - 1} \right \rfloor + \left \lfloor \frac{2^3 m + a_3}{2^n - 1} \right \rfloor + \dots + \left \lfloor \frac{2^{n - 1} m + a_{n - 1}}{2^n - 1} \right \rfloor = m \]holds for all $m \in \mathbb{Z}$. One of the most interesting NT problems I've seen in a while. The main difficulty of this problem is to find the equality case, which can be guessed by doing the case $1 \le n \le 4$. We claim that there's only one $(n - 1)$-tuples that satisfies the above equation, where $a_i = 2^{n - 1} + 2^{i - 1} - 1$ for all $1 \le i \le n - 1$. Sufficiency. We want to prove that \[ \left \lfloor \frac{m}{2^n - 1} \right \rfloor + \sum_{i = 1}^{n - 1} \left \lfloor \frac{2^i m + 2^{n - 1} + 2^{i - 1} - 1}{2^n - 1} \right \rfloor = m\]for all $m \in \mathbb{Z}$. We'll first start by proving the following claims. Claim 01. $\left \lfloor \frac{ax + b}{ac} \right \rfloor = \left \lfloor \frac{ax + b - k}{ac} \right \rfloor$, where $k$ is the smallest nonnegative integer such that $a \mid b - k$. Proof. Let us suppose that $\left \lfloor \frac{ax + b}{ac} \right \rfloor = N = \frac{ax + b - \ell}{ac}$, for some $\ell \in \mathbb{N}_0$. Then, note that we must have $a \mid ax + b - \ell \implies a \mid b - \ell$. By our definition of $k$, we must have $k \le \ell$, and therefore \[ \frac{ax + b}{ac} \ge \frac{ax + b - k}{ac} \ge \frac{ax + b - \ell}{ac} = N \implies N = \left \lfloor \frac{ax + b}{ac} \right \rfloor \ge \left \lfloor \frac{ax + b - k}{ac} \right \rfloor \ge \lfloor N \rfloor = N \]We therefore conclude that $\left \lfloor \frac{ax + b - k}{ac} \right \rfloor = N = \left \lfloor \frac{ax + b}{ac} \right \rfloor$. Claim 02. Let us define \[ A_i = \{ 2^{n - i - 1}(2k + 1) : 0 \le k \le 2^{i - 1} - 1 \} \ \text{and} \ B_i = \{ 2^{n - i - 1}(2k + 1) - 1 : 2^{i - 1} \le k \le 2^i - 1 \} \]then $$\bigcup_{1 \le i \le n - 1} (A_i \cup B_i) = \{ 1, 2, \dots, 2^{n} - 2 \}$$Proof. It's quite clear that $A_i \cap A_j = \varnothing$ and $B_i \cap B_j = \varnothing$ for any $i \not= j$. We'll first prove that $A_i \cap B_j = \varnothing$ for any $1 \le i , j \le n - 1$. Indeed, note that suppose that $x \in A_i \cap B_j$ for some $i,j$, then \[ 2^{n - i - 1}(2k + 1) = 2^{n - j - 1}(2 \ell + 1) - 1 \]for some $0 \le k \le 2^{i - 1} - 1$ and $2^{j - 1} \le \ell \le 2^j - 1$. We either have $i = n - 1$ or $j = n - 1$ by parity. If $i = n - 1$, then $k = 2^{n - j - 2}(2 \ell + 1) - 1 \ge 2^{n - j - 2}(2^j + 1) - 1 \ge 2^{n - 2} > 2^{n - 2} - 1 = 2^{i - 1} - 1$, a contradiction. If $j = n - 1$, then $\ell = 2^{n - i - 2} (2k + 1) \le 2^{n - i - 2}(2^i - 1) < 2^{n - 2} = 2^{j - 1}$, a contradiction. We therefore conclude $A_i \cap B_j = \varnothing$ for any $1 \le i,j \le n - 1$. Now, note that if $x \in A_i$, then $1 \le x = 2^{n - i - 1}(2k + 1) \le 2^{n - i - 1}(2^i - 1) < 2^{n - 1} $. Note that if $y \in B_j$, then $1 \le y = 2^{n - j - 1}(2\ell + 1) - 1 \le 2^{n - j - 1}(2^{j + 1} - 1) - 1\le 2^{n} - 2$. However, we know that \[ \left| \bigcup_{1 \le i \le n - 1} (A_i \cup B_i) \right| = \sum_{i = 1}^{n - 1} |A_i| + |B_i| = \sum_{i = 1}^{n - 1} 2^i = 2^n - 2 \]and therefore the conclusion holds. We'll now prove our claim above. Note that by Hermite's Identity, we have \begin{align*} \left \lfloor \frac{m}{2^n - 1} \right \rfloor + \sum_{i = 1}^{n - 1} \left \lfloor \frac{2^i m + 2^{n - 1} + 2^{i - 1} - 1}{2^n - 1} \right \rfloor &= \left \lfloor \frac{m}{2^n - 1} \right \rfloor + \sum_{i = 1}^{n - 1} \left \lfloor \frac{2^i(m + 2^{n - i - 1} + \frac{2^{i - 1} - 1}{2^i})}{2^n - 1} \right \rfloor \\ &= \left \lfloor \frac{m}{2^n - 1} \right \rfloor + \sum_{i = 1}^{n - 1} \sum_{k = 0}^{2^i - 1} \left \lfloor \frac{m + 2^{n - i - 1} + \frac{2^{i - 1} - 1}{2^i}}{2^n - 1} + \frac{k}{2^i} \right \rfloor \\ &= \left \lfloor \frac{m}{2^n - 1} \right \rfloor + \sum_{i = 1}^{n - 1} \sum_{k = 0}^{2^i - 1} \left \lfloor \frac{2^i m + 2^{n - 1} + 2^{i - 1} - 1 + (2^n - 1)k}{2^i(2^n - 1)} \right \rfloor \\ &= \left \lfloor \frac{m}{2^n - 1} \right \rfloor + \sum_{i = 1}^{n - 1} \sum_{k = 0}^{2^i - 1} \left \lfloor \frac{2^{i}(m + 2^{n - i - 1} + 2^{n - i}k ) + 2^{i - 1} - k - 1 }{2^i(2^n - 1)} \right \rfloor \\ &\stackrel{\textbf{\text{Claim 01}}}{=} \left \lfloor \frac{m}{2^n - 1} \right \rfloor + \sum_{i = 1}^{n - 1} \left( \sum_{k = 0}^{2^{i - 1} - 1} \left \lfloor \frac{m + 2^{n - i - 1} + 2^{n - i} k}{2^n - 1} \right \rfloor + \sum_{k = 2^{i - 1}}^{2^{i} - 1} \left \lfloor \frac{m + 2^{n - i - 1} + 2^{n - i} k - 1}{2^n - 1} \right \rfloor \right) \\ &= \left \lfloor \frac{m}{2^n - 1} \right \rfloor + \sum_{i = 1}^{n - 1} \left( \sum_{k = 0}^{2^{i - 1} - 1} \left \lfloor \frac{m + 2^{n - i - 1}(2k + 1)}{2^n - 1} \right \rfloor + \sum_{k = 2^{i - 1}}^{2^{i} - 1} \left \lfloor \frac{m + 2^{n - i - 1} (2k + 1) - 1}{2^n - 1} \right \rfloor \right) \\ &\stackrel{\textbf{\text{Claim 02}}}{=} \left \lfloor \frac{m}{2^n - 1} \right \rfloor + \sum_{j = 1}^{2^n - 2} \left \lfloor \frac{m + j}{2^n - 1} \right \rfloor \\ &= \sum_{j = 0}^{2^n - 2} \left \lfloor \frac{m + j}{2^n - 1} \right \rfloor \\ &= \left \lfloor (2^n - 1) \cdot \frac{m}{2^n - 1} \right \rfloor = m \end{align*}for all $m \in \mathbb{Z}$ and hence our choice of $a_i$s work. Necessity. We'll now prove that there are no other possible choices of $(a_1, a_2, \dots, a_{n - 1})$. Claim 03. $2^{n - 1} \le a_1, a_2, \dots, a_{n - 1} < 2^n - 1$. Proof. Indeed, for $m = 0$, and since $a_1, a_2, \dots, a_{n - 1} \ge 0$, we have \[ 0 = \sum_{i = 1}^{n - 1} \left \lfloor \frac{a_i}{2^n - 1} \right \rfloor \implies a_1, a_2, \dots, a_{n - 1} < 2^n - 1\]Now, take $m = -2^k$ for some $0 \le k \le n - 1$. We have \begin{align*} -2^k &= \left \lfloor \frac{-2^k}{2^n - 1} \right \rfloor + \sum_{i = 1}^{n - 1} \left \lfloor \frac{a_i - 2^{i + k}}{2^n - 1} \right \rfloor \\ &= -1 + \sum_{i = 1}^{n - k - 1} \left \lfloor \frac{a_i - 2^{i + k}}{2^n - 1} \right \rfloor + \sum_{j = n - k}^{n - 1} \left \lfloor \frac{a_j - 2^{j + k}}{2^n - 1} \right \rfloor \\ &= -1 + \sum_{i = 1}^{n - k - 1} \left \lfloor \frac{a_i - 2^{i + k}}{2^n - 1} \right \rfloor + \sum_{j = n - k}^{n - 1} \left \lfloor \frac{a_j - 2^{j + k - n}}{2^n - 1} \right \rfloor - 2^{j + k - n} \\ &= -1 + \sum_{\ell = 0}^{k - 1} -2^{\ell} + \sum_{i = 1}^{n - k - 1} \left \lfloor \frac{a_i - 2^{i + k}}{2^n - 1} \right \rfloor + \sum_{j = n - k}^{n - 1} \left \lfloor \frac{a_j - 2^{j + k - n}}{2^n - 1} \right \rfloor \\ &= -2^k + \sum_{i = 1}^{n - k - 1} \left \lfloor \frac{a_i - 2^{i + k}}{2^n - 1} \right \rfloor + \sum_{j = n - k}^{n - 1} \left \lfloor \frac{a_j - 2^{j + k - n}}{2^n - 1} \right \rfloor \end{align*}We therefore conclude that for all $0 \le k \le n - 1$, that \[ \sum_{i = 1}^{n - k - 1} \left \lfloor \frac{a_i - 2^{i + k}}{2^n - 1} \right \rfloor + \sum_{j = n - k}^{n - 1} \left \lfloor \frac{a_j - 2^{j + k - n}}{2^n - 1} \right \rfloor = 0 \]As $a_i < 2^n - 1$ for all $1 \le i \le n - 1$, then each floor term can't be positive -- and therefore each has to be exactly $0$. This implies that for all $1 \le i \le n - k - 1$, then $a_i \ge 2^{i + k} $. Fix $i$, vary $k$, then we conclude that $a_i \ge 2^{n - 1}$ for all $1 \le i \le n - 1$. Using Claim 03, we will now prove that $a_k \le 2^{n - 1} + 2^{k - 1} - 1$ for $1 \le k \le n - 1$. First of all, we will prove that $a_{n - 1} \le 2^{n - 1} + 2^{n - 2} - 1$. Take $m = -2^{n - 1} - 1$: we have \begin{align*} -2^{n - 1} - 1 &= -1 + \sum_{i = 1}^{n - 1} \left \lfloor \frac{-2^{i + n - 1} - 2^i + a_i}{2^{n} - 1} \right \rfloor = -1 + \sum_{i = 1}^{n - 1} \left( -2^{i - 1} + \left \lfloor \frac{a_i - 2^i - 2^{i - 1}}{2^n - 1} \right \rfloor \right) \\ -1 &= \sum_{i = 1}^{n - 1} \left \lfloor \frac{a_i - 2^i - 2^{i - 1}}{2^n - 1} \right \rfloor \end{align*}Note that $a_k > 2^{n - 1} \ge 2 \cdot 2^k > 2^k + 2^{k - 1}$ for all $k < n - 1$. Therefore, $a_k - 2^k - 2^{k - 1} > 0$ for $1 \le k \le n - 2$. Therefore, we must have $\left \lfloor \frac{a_{n - 1} - 2^{n - 1} - 2^{n - 2}}{2^n - 1} \right \rfloor = -1$, and the other terms being $0$ since $a_{n - 1} - 2^{n - 1} - 2^{n - 2} \ge - 2^{n - 2} > -(2^n - 1)$. This therefore implies $a_{n - 1} - 2^{n - 1} - 2^{n - 2} < 0$, or $a_{n - 1} \le 2^{n - 1} + 2^{n - 2} - 1$. Now, take $m = -2^{n - 1} - 2^{n - k - 1}$, we have \begin{align*} -2^{n - 1} - 2^{n - k - 1} &= -1 + \sum_{i = 1}^{n - 1} \left \lfloor \frac{-2^{i + n - 1} - 2^{n + i - k - 1} + a_i }{2^n - 1} \right \rfloor \\ &= -1 + \sum_{i = 1}^{n - 1} \left( - 2^{i - 1} + \left \lfloor \frac{-2^{i - 1} - 2^{n + i - k - 1} + a_i}{2^n - 1} \right \rfloor \right)\\ &= -2^{n - 1} -2^{n - k - 1} +1 + \sum_{i = 1}^{k} \left \lfloor \frac{-2^{i - 1} - 2^{n + i - k - 1} + a_i}{2^n - 1} \right \rfloor + \sum_{i = k + 1}^{n - 1} \left \lfloor \frac{-2^{i - 1} - 2^{i - k - 1} + a_i}{2^n - 1} \right \rfloor \\ -1 &= \sum_{i = 1}^{k} \left \lfloor \frac{-2^{i - 1} - 2^{n + i - k - 1} + a_i}{2^n - 1} \right \rfloor + \sum_{i = k + 1}^{n - 1} \left \lfloor \frac{-2^{i - 1} - 2^{i - k - 1} + a_i}{2^n - 1} \right \rfloor \end{align*}We could deduce that $a_i > 2^{n - 1} \ge 2 \cdot 2^{i - 1} > 2^{i - 1} + 2^{i - k - 1}$ for all $i \ge k + 1$, and $a_i \ge 2^{n - 1} > 2^{n - 2} + 2^{i - 1} \ge 2^{n + i - k - 1} + 2^{i - 1} $ for all $i < k$. This therefore implies \[ \left \lfloor \frac{-2^{k - 1} - 2^{n - 1} + a_k}{2^n - 1} \right \rfloor = -1 \]which implies $a_k \le 2^{n - 1} + 2^{k - 1} - 1$ for any $1 \le k \le n- 2$. We will now prove by downward induction that $a_i = 2^{n - 1} + 2^{i - 1} - 1$ for $1 \le i \le n - 1$. Take $m = 2^{n - 1}$: we have \begin{align*} 2^{n - 1} &= \sum_{i = 1}^{n - 1} \left \lfloor \frac{2^{n + i - 1} + a_i}{2^{n} - 1} \right \rfloor = \sum_{i = 1}^{n - 1} \left( 2^{i - 1} + \left \lfloor \frac{2^{i - 1} + a_i}{2^{n} - 1} \right \rfloor \right) \\ 1 &= \sum_{i = 1}^{n - 1} \left \lfloor \frac{2^{i - 1} + a_i}{2^{n} - 1} \right \rfloor \end{align*}Note that $a_i + 2^{i - 1} \le 2^{n - 1} + 2^{i - 2} + 2^{i - 1} - 1 < 2^{n - 1} + 2^{i} - 1 < 2^n - 1 $ for all $i \le n - 2$. This therefore forces \[ \left \lfloor \frac{a_{n - 1} + 2^{n - 2}}{2^n - 1} \right \rfloor = 1 \implies a_{n - 1} \ge 2^n - 2^{n - 2} + 1 = 2^{n - 1} + 2^{n - 2} + 1 \]We conclude that $a_{n - 1} = 2^{n - 1} + 2^{n- 2} + 1$. Now, take $m =-2^{n - 1} + 2^{n - k - 1}$. We have \begin{align*} -2^{n - 1} + 2^{n - k - 1} &= -1 + \sum_{i = 1}^{n - 1} \left \lfloor \frac{-2^{i + n - 1} + 2^{n + i - k - 1} + a_i}{2^n - 1} \right \rfloor \\ &= -1 + \sum_{i = 1}^{n - 1} \left( -2^{i - 1} + \left \lfloor \frac{-2^{i - 1} + 2^{n + i - k - 1} + a_i}{2^n - 1} \right \rfloor \right) \\ &= -2^{n - 1} + \sum_{i = 1}^{k} \left \lfloor \frac{-2^{i - 1} + 2^{n + i - k - 1} + a_i}{2^n - 1} \right \rfloor + \sum_{i = k + 1}^{n - 1} \left( 2^{i - k - 1} + \left \lfloor \frac{-2^{i - 1} + 2^{i - k - 1} + a_i}{2^n - 1} \right \rfloor \right) \\ &= -2^{n - 1} + 2^{n - k - 1} - 1 + \sum_{i = 1}^{k} \left \lfloor \frac{-2^{i - 1} + 2^{n + i - k - 1} + a_i}{2^n - 1} \right \rfloor + \sum_{i = k + 1}^{n - 1} \left \lfloor \frac{-2^{i - 1} + 2^{i - k - 1} + a_i}{2^n - 1} \right \rfloor \\ 1 &= \sum_{i = 1}^{k} \left \lfloor \frac{-2^{i - 1} + 2^{n + i - k - 1} + a_i}{2^n - 1} \right \rfloor + \sum_{i = k + 1}^{n - 1} \left \lfloor \frac{-2^{i - 1} + 2^{i - k - 1} + a_i}{2^n - 1} \right \rfloor \end{align*}Note that $a_i + 2^{n + i - k - 1} - 2^{i - 1} \le 2^{n - 1} + 2^{i - 1} + 2^{n - 2} - 2^{i - 1} - 1 < 2^n - 1$ for $1 \le i \le k - 1$, and therefore \[ \left \lfloor \frac{-2^{i - 1} + 2^{n + i - k - 1} + a_i}{2^n - 1} \right \rfloor = 0 \ \text{for } 1 \le i \le k - 1 \]However, by induction hypothesis, $0 < a_i + 2^{i - k - 1} - 2^{i - 1} = 2^{n - 1} + 2^{i - k - 1} < 2^n - 1$ for all $i \ge k + 1$, and therefore \[ \left \lfloor \frac{-2^{i - 1} + 2^{i - k - 1} + a_i}{2^n - 1} \right \rfloor = 0 \]which therefore implies \[ \left \lfloor \frac{-2^{k - 1} + 2^{n - 1} + a_k}{2^n- 1} \right \rfloor = 1 \implies a_k \ge 2^{n - 1} + 2^{k - 1} + 1 \]Combine with the bound we get $a_k \le 2^{n - 1} + 2^{k - 1} + 1$, we conclude that $a_k = 2^{n - 1} + 2^{k - 1} + 1$ for all $1 \le k \le n - 1$, which is what we wanted.

@Above What a beatiful solution! Again, in my opinion this question was really hard and it was a bit absurd to give it on TST, the solution I gave below is from one of the official solutions: (Note that there can be latex errors)