5. If for any positive divisor $a$ of a positive integer $n$ we have $(a+1) \mid (n+1)$, then $n$ is called a "good number". Find all good numbers.
5. Good numbers are 1 and all odd primes.
Let $n$ be a good number.
Since $1 \mid n$, and $(1+1) \mid (n+1)$, therefore, $n$ is an odd number.
Obviously, any odd prime is a good number.
If there is an odd composite number $n = ab$ that is a good number, without loss of generality, assume $3 \leqslant b \leqslant a$.
Then $(a+1) \mid (n+1)$.
Thus, $b-1 = (n+b) - (n+1)$
$$
= b(a+1) - (n+1)
$$

is a multiple of $a+1$.
Also, $b-1 \geqslant 2$, so $b-1 \geqslant a+1$. This is a contradiction.