9.2. Let $n$ be a natural number. An integer $a>2$ is called $n$-decomposable if $a^{n}-2^{n}$ is divisible by each number of the form $a^{d}+2^{d}$, where $d$ is a natural divisor of $n$, different from $n$. Find all composite natural numbers $n$ for which there exists an $n$-decomposable number.

(S. Kudrya)
Answer. $n=2^{m}$ for $m>1$.

Solution. For $n=2^{m}$, any natural number $a>2$ is $n$-decomposable due to the equality

$$
a^{2^{m}}-2^{2^{m}}=(a-2)\left(a^{1}+2^{1}\right)\left(a^{2}+2^{2}\right) \ldots\left(a^{2^{m-1}}+2^{2^{m-1}}\right)
$$

Indeed, among the factors on the right side, there are all numbers of the form $a^{d}+2^{d}$, where $d$ is a divisor of $n$, less than $n$.

Now suppose $n$ is not a power of two, then it has an odd prime divisor $p$ and $n=p k$, where $k>1$ is a natural number. Assume there exists an $n$-decomposable number $a$. Then $a^{p k}-2^{p k}$ is divisible by $a^{k}+2^{k}$. Moreover, $a^{p k}+2^{p k}$ is also divisible by $a^{k}+2^{k}$, since $p$ is odd. Therefore, $2 \cdot 2^{p k}=\left(a^{p k}+2^{p k}\right)-\left(a^{p k}-2^{p k}\right)$ is also divisible by $a^{k}+2^{k}$. Thus, the number $a^{k}+2^{k}>1$ is a divisor of a power of two and, therefore, is itself a power of two.

In particular, this implies that $a$ is even, i.e., $a=2 b$ for some natural number $b>1$. Then $b^{k}+1=\frac{a^{k}+2^{k}}{2^{k}}>1$ is also a power of two and, therefore, $b$ is odd. If $k$ is even, then the number $b^{k}+1$ gives a remainder of 2 when divided by 4, and this number is greater than 2, which is impossible. Therefore, $k$ must be odd. But then the number

$$
b^{k}+1=(b+1)\left(b^{k-1}-b^{k-2}+b^{k-3}-\ldots-b+1\right)
$$

cannot be a power of two, since the second parenthesis is odd and greater than one. Thus, an $n$-decomposable number does not exist.