$:$ Senororov B.A.

Given the equality $\left(a^{m_{1}}-1\right) \ldots\left(a^{m_{n}}-1\right)=\left(a^{k_{1}}+1\right) \ldots\left(a^{k_{l}}+1\right)$, where $a, n, l$ and all exponents are natural numbers, and $a>1$.

Find all possible values of the number $a$.
Suppose that for some $a>3$ we have $A=\left(a^{m_{1}}-1\right) \ldots\left(a^{m_{n}}-1\right)=\left(a^{\left(k_{1}+1\right) \ldots(} a^{k_{l}}+1\right)$. First, we will prove that any $m_{i}$ and the number $a-1$ are powers of two. Let $\gamma$ be an odd divisor of the number $m_{i}$, and let $p$ be any prime divisor of the number $a^{\gamma}-1$. Then in the right product, there will be a factor $a^{k_{j}}+1$ that is divisible by $p$. Therefore, the number $\left(a^{\gamma k_{j}}+1\right)-\left(a^{\gamma k_{j}}-1\right)=2$ is divisible by $p$, meaning all prime divisors of the number $a^{\gamma}-1$ are twos, and $\left(a^{\gamma-1}+\ldots+a+1\right)(a-1)=a^{\gamma}-1=2^{s}$. Since $a>3$, then $s>1$ and $\gamma$ is odd, so the expression in the first parenthesis is an odd divisor of the number $2^{s}$. Thus, $\gamma$ and $a-1=2^{s}$. In particular, $a-1$ is divisible by 4, so $a^{k}+1 \equiv 2(\bmod 4)$.

Each number in the left part of the original equation can be represented as $a^{2^{d}}-1=(a-1)(a+1)\left(a^{2}+1\right)\left(a^{4}+1\right) \ldots \left(a^{2^{d-1}}+1\right)=2^{s \cdot} 2^{d} a_{0} \ldots a_{d-1}$, where

$a_{i}=1 / 2\left(a^{2^{i}}+1\right)$ are odd numbers. Note that for $m>n$, the number $a^{2^{m}}-1$ is divisible by $a^{2^{n}}+1$, so the GCD $\left(a_{m}, a_{n}\right) \leq \left(a^{2^{m}}+1\right)-\left(a^{2^{m}}-1\right)=2$; since $a_{i}$ are odd, they are pairwise coprime. Let $q$ be the maximum power of two that divides $m_{i}$ or $k_{j}$. Then $A=2^{N} a_{0}^{N_{0}} \ldots a_{q}^{N_{q}}$, where $N>N_{0}+\ldots+N_{q}$, as this is true for any representation of $a^{2^{d}}-1$.

No factor $a^{k_{j}}+1$ is divisible by 4, so there are at least $N$ such factors in the representation of the number $A$. But each of them is divisible by one of the numbers $a_{i}$: if $k_{j}=2^{r} b$, where $b$ is odd, then $a^{k_{j}}+1$ is divisible by $a_{r}$. Since $N>N_{0}+\ldots+N_{q}$, some number $a_{i}$ divides more than $N_{i}$ numbers $a^{k_{j}}+1$, and therefore $A$ is divisible by a higher power of $a_{i}$ than $N_{i}$. This contradicts the fact that $a_{i}$ are odd and pairwise coprime.

## Answer

2 and 3