Consider the set: $ A = \{e_1a_1 + e_2a_2 + ... + e_na_n: 0 \leq e_1, e_2, ..., e_n \leq m - 1\}$. It clearly consists of $ m^n$ elements. If some two of them would be congruent mod $ m^n$, i.e. $ e_1a_1 + ... + e_na_n \equiv e'_1a_1 + ... + e'_na_n \pmod{m^n}$ then it would follow that $ (e_1 - e'_1)a_1 + ... + (e_n - e'_n)a_n \equiv 0 \mod{m^n}$ and the numbers $ e_1 - e'_1$, ..., $ e_n - e'_n$ satisfy given conditions. So suppose that all numbers from the set $ A$ give different residues, which means that they give all residues. Consider $ \omega$ the primitive root of $ 1$ of degree $ m^n$-th. Since $ 1 + \omega^{1} + \omega^{2} + ... + \omega^{m^{n} - 1} = 0$ we also have $ \sum_{k \in A} \omega^k = 0$. But on the other hand it's not hard to see that $ \sum_{k \in A} \omega^k = \prod_{i = 1}^{n} (1 + \omega^{a_i} + \omega^{2a_i} + ... + \omega^{(m - 1)a_i}).$ So it follows that for some $ i$: $ 0 = 1 + \omega^{a_i} + \omega^{2a_i} + ... + \omega^{(m - 1)a_i} = \frac{\omega^{ma_i} - 1}{\omega - 1}$ and it's a contradiction since $ m^{n - 1}$ doesn't divide $ a_i$.

Let $ B$ be the set of all $ n$-tuples $ b=(b_1, b_2, \dots, b_n)$, where each $ b_i$ is an integer with $ 0\le i<m$. For $ b\in B$, define $ f(b)=b_1a_1+b_2a_2+\dots+b_na_n$. As in the above solution, suppose $ f(b)$ gives a complete residue class. Then \[ (0+1+2+\cdots+m-1)(a_1+a_2+\cdots+a_n)=\displaystyle\sum_{b\in B}{f(b)}\equiv 0+1+2+\cdots+m^{n}-1\pmod {m^n}\] We have \[ (a_1+a_2+\cdots+a_n)\dfrac{m(m-1)}{2}\equiv \dfrac{m^{n}(m^{n}-1)}{2}\pmod {m^n}\] which means $ (a_1+a_2+\cdots+a_n)\equiv 0\pmod {m^{n-1}}$. Now by the assumption there exist a tuple $ b$ such that $ f(b)\equiv -(a_1+a_2+\cdots+a_n)\pmod {m^n}$. Taking $ e_i=b_i+1-m$ finishes the problem.

nealth wrote: Then \[ (0 + 1 + 2 + \cdots + m - 1)(a_1 + a_2 + \cdots + a_n) = \displaystyle\sum_{b\in B}{f(b)}\equiv 0 + 1 + 2 + \cdots + m^{n} - 1\pmod {m^n} \] Why was this? Additionally, where did you use the that ai are not multiples of $ m^{n-1}$? Notice that the problem is not true without this condition.

Imo short list 2002 n6 Complex combinatorics