Problem 1046

Solution. Suppose that $n=p_{1}^{s_{1}} p_{2}^{s_{2}} \ldots p_{k}^{s_{k}}$, where $p_{1}, \ldots, p_{k}$ are distinct primes and $s_{i} \geqslant 1$ for each $i$, and that the sum of all positive divisors of $n$, which is given by

$$\left(1+p_{1}+p_{1}^{2}+\cdots+p_{1}^{s_{1}}\right)\left(1+p_{2}+p_{2}^{2}+\cdots+p_{2}^{s_{2}}\right) \ldots\left(1+p_{k}+p_{k}^{2}+\cdots+p_{k}^{s_{k}}\right)$$

is a perfect power of 2 . Then each of the factors

$$f_{i}=1+p_{i}+p_{i}^{2}+\cdots+p_{i}^{s_{i}}$$

is also a perfect power of 2 greater than 1 and hence both $p_{i}$ and $s_{i}$ are odd. Suppose that $s_{i}>1$. In this case we have

$$f_{i}=\left(1+p_{i}\right)\left(1+p_{i}^{2}+p_{i}^{4}+\cdots+p_{i}^{s_{i}-1}\right)$$

Since $f_{i}$ has no odd divisor greater than 1 , the even integer $s_{i}-1$ (which is supposed to be positive) must be of the form $4 k+2$ and thus we can make another factorization

$$f_{i}=\left(1+p_{i}\right)\left(1+p_{i}^{2}\right)\left(1+p_{i}^{4}+p_{i}^{8}+\cdots+p_{i}^{s_{i}-3}\right)$$

Consequently, both $1+p_{i}$ and $1+p_{i}^{2}$ are powers of 2 , hence $1+p_{i} \mid 1+p_{i}^{2}$, which contradicts to $1+p_{i}^{2}=\left(1+p_{i}\right)\left(p_{i}-1\right)+2$ (as $1+p_{i} \mid 2$ is impossible). This means that $s_{i}=1$ for each $i$ and thus the number of divisors of $n$ equals $2^{k}$.

Note that the above solution can be finished without observing the fact that $1+p_{i}$ and $1+p_{i}^{2}$ cannot be powers of 2 at the same time. Indeed, repeating the procedure of factorization we get finally

$$f_{i}=\left(1+p_{i}\right)\left(1+p_{i}^{2}\right)\left(1+p_{i}^{4}\right) \ldots\left(1+p_{i}^{2_{i}}\right)$$

hence $s_{i}=2^{t_{i}+1}-1$ with some $t_{i} \geqslant 0$ for each $i$ and thus the number of divisors of $n$ equals $2^{k+t_{1}+t_{2}+\cdots+t_{k}}$. (As we know from the original solution, $t_{i}=0$ for each $i$.)