bump this I couldn't solve this problem in the test TT But I think this is an interesting problem

Let $f(x) = x^{b_1}(x^0 + x^{b_2} + \cdots x^{(m-1)b_2}) \cdots (x^0 + x^{b_k} + \cdots x^{(m-1)b_k})$ and $g(x) = (x^0 + x^{b_1} + \cdots x^{(m-1)b_1})x^{2b_2} (x^0 + x^{b_3} + \cdots x^{(m-1)b_3}) \cdots (x^0 + x^{b_k} + \cdots x^{(m-1)b_k})$ Also, let $\omega \neq 1$ as a root of the equation $\omega^n =1$ It is easy to see that the number of elements such that $a_1 = 1$ is $\frac{f(\omega^0) + f(\omega^1) + \cdots + f(\omega^{n-1})}{n}$ and the number of elements such that $a_2 = 2$ is $\frac{g(\omega^0) + g(\omega^1) + \cdots + g(\omega^{n-1})}{n}$ It is trivial that $f(\omega^0) = g(\omega^0)$ Hence, it is enough to show that $| \sum_{l=1}^{n-1}{\left( f(\omega^l) - g(\omega^l) \right)} | = 0 \text{ or } n$ $f(\omega^l) = \frac{\omega^{lb_1}}{\omega^0 + \omega^{b_1l} + \cdots + \omega^{(m-1)b_1l}} \times \prod_{i=1}^{k}{\frac{\omega^{mlb_i} - 1}{\omega^{lb_i}-1}}$ Here, $\{ mb_1l, mb_2l ,mb_3l , \cdots mb_kl \} \equiv \{ lb_1, lb_2, \cdots lb_k \}$ in $\mod {n}$ Hence, $f(\omega^l) = \frac{\omega^{lb_1}}{\omega^0 + \omega^{b_1l} + \cdots + \omega^{(m-1)b_1l}} = \omega^l \times (\omega^l - 1) \times \frac{1}{\omega^{ml} - 1}$ In the same way, $g(\omega^l) = \omega^{2b_2l} \times (\omega^{b_2l} - 1) \times \frac{1}{\omega^{mb_2l}-1}$

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Let's calculate $T = \sum_{l=1}^{n-1}{\omega^l \times (\omega^l - 1) \times \frac{1}{\omega^{ml} - 1}} - \sum_{l'=1}^{n-1}{\omega^{2b_2l'} \times (\omega^{b_2l'} - 1) \times \frac{1}{\omega^{mb_2l'}-1}}$ $T = \sum_{l=1}^{n-1}{\omega^l \times (\omega^l - 1) \times \frac{1}{\omega^{ml} - 1} - \omega^{2l} \times (\omega^{l} - 1) \times \frac{1}{\omega^{ml}-1}}$ $\qquad = \sum_{l=1}^{n-1}{(\omega^l - \omega^{2l}) \times \frac{\omega^l-1}{\omega^{ml} - 1}}$ Since $(m,n) = 1$, $m' < n$ such that $mm' \equiv 1 \pmod n$ exists Also, $m' \in \mathbb{Z}_n^{\times}$ so $\{ 1,2, \cdots n-1 \} \equiv \{ m', 2m' \cdots (n-1)m' \}$ So $T = \sum_{l=1}^{n-1}{(\omega^{m'l} - \omega^{2m'l}) \times \frac{\omega^{m'l}-1}{\omega^{l} - 1}}$ $\qquad \qquad = \sum_{l=1}^{n-1}{(\omega^{m'l} - \omega^{2m'l}) \times (\omega^0 + \omega^{l} + \cdots \omega^{(m'-1)l})}$ Let $P(x) = x^{m'} + x^{m'+1} + \cdots x^{2m' - 1} - x^{2m'} - x^{2m'+1} - \cdots - x^{3m'-1}$ Then $T = \sum_{l=1}^{n-1}{P(\omega^l)}$ If $n \ge 3m'$ $T = 0$ If $3m' > n \ge 2m'$ $T = -n$ If $2m' > n \ge m'$ $T = n \text{ or } 0$ Hence, $|T| = 0 \text{or} n$ Then $| \sum_{l=1}^{n-1}{\left( f(\omega^l) - g(\omega^l) \right)} | = 0 \text{ or } n$ Therefore, the difference of number of elements such that $a_1 = 1$ and $a_2 = 2$ is maximum $1$

Seungjun_Lee wrote: $n$ is a natural number larger than $3$ and denote all positive coprime numbers with $n$ as $1= b_1 < b_2 < \cdots b_k$. For a positive integer $m$ which is larger than $3$ and is coprime with $n$, let $A$ be the set of tuples $(a_1,a_2, \cdots a_k)$ satisfying the condition. $$\textbf{Condition}: \text{For all integers } a_i, 0 \le a_i < m \text{ and } a_1b_1 + a_2b_2 + \cdots a_kb_k \text{ is a mutiple of } n$$For elements of $A$, show that the difference of number of elements such that $a_1 = 1$ and the number of elements such that $a_2 = 2$ maximum $1$ maybe you should say that "denote all positive coprime numbers $t$ with $n$ satisfying $t \le n$