2. Given 1995 complex numbers. If we multiply any two of them (not necessarily distinct), we get one of these numbers again. Find all sets of numbers that have this property.
2. All numbers by modulus must be equal to 1 (otherwise, multiplying that number by itself would yield infinitely many numbers). (points)

Let $\alpha_{k}$ be the argument of the $k$-th number, $z_{k}=\cos \alpha_{k}+i \sin \alpha_{k}$. When multiplying, the arguments add up.

1) All numbers $\alpha_{k}$ are of the form $\frac{p_{k}}{q_{k}} \cdot 2 \pi$, where $p_{k}$ and $q_{k}$ are natural numbers (otherwise, multiplying the number by itself would yield infinitely many numbers).
2) Let $q_{\max }$ be the largest number in the set $Q=\left\{q_{1}, \ldots, q_{1995}\right\}$. Suppose some $q_{k}$ does not divide $q_{\max }$. Then $\arg z_{\max } z_{k}=\alpha_{\max }+\alpha_{k}=\left(\frac{p_{\max }}{q_{\max }}+\frac{p_{k}}{q_{k}}\right) \cdot 2 \pi=\frac{p^{\prime}}{q^{\prime}} \cdot 2 \pi, q^{\prime}=\text{gcd}\left(q_{\max }, q_{k}\right)$. Then $q^{\prime}>q_{\max }$, which contradicts the assumption. Therefore, every $q_{k}$ divides $q_{\max }$. (5 points)
3) Each argument is of the form $\frac{p_{k}}{q_{\max }}$. Let $p_{1}>0$ be the smallest among the numbers $\left\{p_{k}\right\}$. The numbers $p_{1}$ and $q_{\max }$ are relatively prime. Then the numbers $\frac{p_{1}}{q_{\max }}, \frac{2 p_{1}}{q_{\max }}, \ldots, \frac{1995 p_{1}}{q_{\max }}$ are all the arguments of the given complex numbers. Let $\frac{q_{\max }}{k p_{1}} \equiv r_{k}\left(\bmod q_{\max }\right)$. Then for $1 \leq k<j \leq 1995$ it would be $(j-k) p_{1} \equiv 0\left(\bmod q_{\max }\right)$, i.e., $q_{\max }$ would divide $j-k$, which is impossible. Therefore, it must be $q_{\max } \geq 1995$.

(5 points)

4) It must be $q_{\max }=1995$ (otherwise, we would have fewer than 1995 numbers, since all possible arguments are of the form $\alpha_{k}=\frac{k}{q_{\max }} \cdot 2 \pi, \quad k=0,1,2, \ldots, q_{\max }$).

Finally, $\alpha_{k}=\frac{k}{1995} \cdot 2 \pi, \quad k=0,1,2, \ldots, 1994$. These numbers are the solutions of the equation $x^{1995}=1$.

(5 points)