Problem 1682

1. **Case 1: $a \neq 0$**

- Given that $f_1(x), f_2(x), \ldots, f_n(x)$ are additive functions, we know that for every $x \in \mathbb{R}$, $f_1(x) f_2(x) \cdots f_n(x) = ax^n$.
- Since $a \neq 0$, none of the $f_k(1)$ are zero. Therefore, $a = \prod_{k=1}^n f_k(1)$.
- Define $g_k(x) = \frac{f_k(x)}{f_k(1)}$. Then, we have:
$$\prod_{k=1}^n g_k(x) = \frac{f_1(x) f_2(x) \cdots f_n(x)}{f_1(1) f_2(1) \cdots f_n(1)} = \frac{ax^n}{a} = x^n$$
- Fix any $\alpha \in \mathbb{R}$. For any rational $p$, we have:
$$\prod_{k=1}^n g_k(p\alpha) = (p\alpha)^n$$
- Both sides of the equation are polynomials in $p$. Since they agree for infinitely many values of $p$, they must be identical as polynomials. Therefore, for each $k$, $g_k(\alpha) = \alpha$ for all $\alpha \in \mathbb{R}$.
- This implies that $f_k(x) = f_k(1)x$ for all $k$ and $x \in \mathbb{R}$. Let $b = f_i(1)$ for some $i$. Then $f_i(x) = bx$.

2. **Case 2: $a = 0$**

- Assume $a = 0$. This means $f_1(x) f_2(x) \cdots f_n(x) = 0$ for all $x \in \mathbb{R}$.
- Suppose there exists some $\alpha_i$ such that $f_i(\alpha_i) \neq 0$. Then, we can find rationals $p_i$ such that:
$$f_k\left(\sum_{i=1}^n p_i \alpha_i\right) = \sum_{i=1}^n p_i f_k(\alpha_i) \neq 0$$
- This would contradict the fact that $\prod_{k=1}^n f_k$ evaluated at $\sum_{i=1}^n p_i \alpha_i$ is zero. Therefore, at least one of the $f_i$'s must be identically zero.
- For that $i$, we can take $b = 0$ and $f_i(x) = 0$.

Conclusion:
In both cases, we have shown that there exists $b \in \mathbb{R}$ and $i \in \{1, 2, \ldots, n\}$ such that $f_i(x) = bx$ for all $x \in \mathbb{R}$.

The final answer is $\boxed{b \in \mathbb{R} \text{ and } i \in \{1, 2, \ldots, n\} \text{ such that } f_i(x) = bx}$