2. Given an integer $a$ greater than 1. Find the arithmetic sequence with the first term $a$, which contains exactly two of the numbers $a^{2}, a^{3}, a^{4}, a^{5}$ and has the largest possible difference. (We do not assume that the difference must be an integer.)
2. The arithmetic sequence $\mathcal{A}$ with the first term $a$ and positive difference $d$ contains exactly those of the numbers $a^{2}, a^{3}, a^{4}, a^{5}$ for which the corresponding difference

$$
\begin{aligned}
a^{2}-a & =a(a-1) \\
a^{3}-a & =a(a-1)(a+1) \\
a^{4}-a & =a(a-1)\left(a^{2}+a+1\right), \\
a^{5}-a & =a(a-1)(a+1)\left(a^{2}+1\right)
\end{aligned}
$$

is an integer multiple of the number $d$. Assume that $\mathcal{A}$ contains exactly two of the numbers $a^{2}, a^{3}, a^{4}, a^{5}$.

If $a^{2} \in \mathcal{A}$, the first difference $a(a-1)$ is an integer multiple of $d$. Since the number $a$ is an integer, the other three differences in (1) (which are clearly integer multiples of the first difference) are also integer multiples of $d$. This means that $\mathcal{A}$ contains all the numbers $a^{2}, a^{3}, a^{4}, a^{5}$, which contradicts the requirements of the problem. Therefore, $a^{2} \notin \mathcal{A}$, so $\mathcal{A}$ contains exactly two of the numbers $a^{3}$, $a^{4}, a^{5}$.

If $a^{3} \notin \mathcal{A}$ or $a^{4}, a^{5} \in \mathcal{A}$, both expressions

$$
\frac{a^{4}-a}{d}, \quad \frac{a^{5}-a}{d}
$$

are integers. Then their combination

$$
\frac{a^{5}-a}{d}-a \cdot \frac{a^{4}-a}{d}=\frac{a^{2}-a}{d}
$$

is also an integer, i.e., $a^{2} \in \mathcal{A}$, which leads to a contradiction as we already know. Therefore, $a^{3} \in \mathcal{A}$.

We have shown that the desired arithmetic sequence $\mathcal{A}$ must contain the number $a^{3}$. For its difference $d$, the estimate $d \leqq a^{3}-a$ holds, and equality occurs (i.e., the difference will be the largest) when $a^{3}$ is its second term. The arithmetic sequence with the first term $a$ and difference $d=a^{3}-a$ certainly does not contain the number $a^{2}$, contains $a^{3}$, and also contains $a+\left(a^{2}+1\right)\left(a^{3}-a\right)=a^{5}$. It remains to verify that it does not contain $a^{4}$. This follows from the fact that the expression

$$
\frac{a^{4}-a}{d}=\frac{a^{4}-a}{a^{3}-a}=\frac{a^{2}+a+1}{a+1}=a+\frac{1}{a+1}
$$

is not an integer for any positive integer $a$.

Conclusion. The desired sequence is an arithmetic sequence with the first term $a$ and difference $d=a^{3}-a$.

Note. If we guess the result and show that the sequence with difference $d=a^{3}-a$ contains $a^{3}, a^{5}$ and does not contain $a^{2}, a^{4}$, it is sufficient to complete the solution by showing only that the difference of any suitable sequence $\mathcal{A}$ cannot be greater than $a^{3}-a$. If it were greater, the sequence $\mathcal{A}$ could not contain $a^{2}$ or $a^{3}$, and would therefore have to contain $a^{4}$ and $a^{5}$. From this, as in the given solution, we can deduce $a^{2} \in \mathcal{A}$, leading to a contradiction.

For a complete solution, award 6 points. If the student correctly derives the estimate $d \leqq a^{3}-a$ and states that the largest possible difference is $a^{3}-a$, but does not show that the sequence with this difference satisfies the conditions, award only 4 points. Award an additional 1 point for verifying that in this case $a^{5} \in \mathcal{A}$, or 1 point for proving that $a^{4} \notin \mathcal{A}$.

If, on the other hand, the student shows that the sequence with difference $a^{3}-a$ contains $a^{3}, a^{5}$ and does not contain $a^{2}, a^{4}$, but does not explain why the difference cannot be greater, award only 3 points.

In the case where the solver does not reach the considerations about the difference $a^{3}-a$, award 1 point for proving that $a^{2} \notin \mathcal{A}$ (i.e., for proving the implication $a^{2} \in \mathcal{A} \Rightarrow a^{3}, a^{4}, a^{5} \in \mathcal{A}$), 1 point for proving the implication $a^{3} \in \mathcal{A} \Rightarrow a^{5} \in \mathcal{A}$, and 3 points for proving the implication $a^{4}, a^{5} \in \mathcal{A} \Rightarrow a^{2} \in \mathcal{A}$; however, in this sequence, award no more than 4 points in total.