10.5. On the board, all proper divisors of some composite natural number $n$, increased by 1, were written down. Find all such numbers $n$ for which the numbers on the board turn out to be all proper divisors of some natural number $m$. (Proper divisors of a natural number $a>1$ are all its natural divisors, different from $a$ and from 1.) (A. Khryabrov)
Answer: $n=4$ or $n=8$.

First solution. Note that the number 2 is not written on the board, because 1 is not a proper divisor of $n$; hence, $m$ is odd. Therefore, all the divisors of $m$ are odd, and thus all the divisors of $n$ are even. Thus, $n$ does not divide by odd prime numbers, meaning $n$ is a power of two (and all its divisors are also powers of two).

If $n$ is divisible by 16, then 4 and 8 are its proper divisors, so 5 and 9 are written on the board. Therefore, $m$ is divisible by 45, and in particular, 15 is its proper divisor. But the number 15 could not have been written, since 14 is not a power of two. Therefore, $n$ cannot be divisible by 16.

The remaining composite powers of two, $n=4$ and $n=8$, fit: for them, we can respectively set $m=9$ and $m=15$.

Second solution. Let $a_{1}n / a_{2}>\ldots>$ $>n / a_{k}$ be the proper divisors of the number $n$. Then they correspond to $a_{k}, a_{k-1}, \ldots, a_{1}$, meaning $a_{i} a_{k+1-i}=n$ for all $i=1,2, \ldots, k$. A similar reasoning can be applied to the divisors of the number $m$.

Suppose $k \geqslant 3$. Then $a_{1} a_{k}=a_{2} a_{k-1}=n$ and $\left(a_{1}+1\right)\left(a_{k}+1\right)=$ $=\left(a_{2}+1\right)\left(a_{k-1}+1\right)=m$, from which $a_{1}+a_{k}=a_{2}+a_{k-1}=m-$ $-n-1$. We see that the pairs of numbers $\left(a_{1}, a_{k}\right)$ and ( $a_{2}, a_{k-1}$ ) have the same sum and product; by Vieta's theorem, they are pairs of roots of the same quadratic equation, meaning these pairs must coincide - a contradiction.

Thus, $k \leqslant 2$; this is possible if $n=p^{2}, n=p^{3}$, or $n=p q$, where $p$ and $q$ are prime numbers (in the last case, we assume $p<q$). Note that $p$ is the smallest proper divisor of $n$, and from the condition, $p+1$ is the smallest proper divisor of $m$, meaning $p+1$ is a prime number. Therefore, $p=2$. The cases $n=2^{2}$ and $n=2^{3}$ fit, as noted in the first solution. The case $n=2 q$ is impossible. Indeed, in this case, $m$ has only two proper divisors, 3 and $q+1$, so $m=3(q+1)$, and $q+1$ is a prime number different from 3, or nine; both cases are impossible for a prime $q$.