2. What is the smallest number of different integers that need to be taken so that among them one can choose both a geometric and an arithmetic progression of length $5$?

(M.A. Evdokimov)
Solution. Let's provide an example of six integers that satisfy the condition: $-8, -2, 1, 4, 10, 16$. The numbers $1, -2, 4, -8, 16$ form a geometric progression, while the numbers $-8, -2, 4, 10, 16$ form an arithmetic progression.

We will show that no five distinct integers satisfy the condition of the problem. Suppose the contrary: let there be five integers that simultaneously form a geometric progression and, possibly in a different order, an arithmetic progression. Then they have the form $b$, $b q, b q^{2}, b q^{3}, b q^{4}$ where $b \in \mathbb{Z}$. Note that $b \neq 0$ and $q \neq 0$ by the definition of a geometric progression. The numbers $b, b q^{2}, b q^{4}$ are always of the same sign and in an arithmetic progression, they either appear consecutively for $q > 0$ or $q < 0$. In any case, the equality $2 b q^{2} = b + b q^{4}$ must hold, i.e., $b\left(q^{2}-1\right)^{2}=0$, from which $q = \pm 1$, but then there are equal numbers among them. Contradiction. Therefore, five numbers are insufficient.

Answer: 6.