All proper divisors of some composite natural number $n$, increased by 1, were written on the board. Find all such numbers $n$ for which the numbers on the board turn out to be all proper divisors of some natural number $m$.

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Note that the number 2 is not written on the board, because 1 is not a proper divisor of $n$; hence, $m$ is odd. This means that all the divisors of $m$ are odd, and therefore all 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).

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 $n=8$ work: for them, we can respectively set $m=$ 9 and $m=15$.

## Answer

$n=4$ or $n=8$.