
2. Find an arithmetic progression with the first element an integer $a>1$ such that exactly two numbers from $a^{2}, a^{3}, a^{4}, a^{5}$ are elements of the progression as well and it's difference is as large as possible (the difference does not have to be an integer).

(Jaromír Šimša)


Solution. An arithmetic progression $\mathcal{A}$ with the first element $a$ and a positive difference $d$ contains those numbers from $a^{2}, a^{3}, a^{4}, a^{5}$ for which the corresponding difference from

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

is an integer multiple of $d$.

Let us suppose $\mathcal{A}$ contains just two numbers out of $a^{2}, a^{3}, a^{4}, a^{5}$. If $a^{2} \in \mathcal{A}$, then $a(a-1)$ is an integer multiple of $d$. Since $a$ is an integer, then all the differences in (1) (all of them are integer multiples of $a(a-1)$ ) are integer multiples of $d$. Thus $\mathcal{A}$ contains all numbers $a^{2}, a^{3}, a^{4}, a^{5}$, therefore the sought $\mathcal{A}$ does not contain $a^{2}$, and it contains exactly two from $a^{3}, a^{4}, a^{5}$.

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

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

are integers.

Then also

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

is integer. Consequently $a^{2} \in \mathcal{A}$, contradiction. Thus $a^{3} \in \mathcal{A}$. Then we have $d \leqslant a^{3}-a$ for the difference $d$, where the equation holds ( $d$ is the largest one) if $a^{3}$ is the second element of the progression. Then $a^{2} \notin \mathcal{A}$, and $a+\left(a^{2}+1\right)\left(a^{3}-a\right)=a^{5} \in \mathcal{A}$. Also

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

is not an integer for any positive integer $a$, which shows $a^{4} \notin \mathcal{A}$.

Conclusion. The sought arithmetic progression is the progression with the first element $a$ and the difference $d=a^{3}-a$.
