5. The positive integer $n$ has exactly 4 positive divisors (including 1 and $n$). It is known that $n+1$ is four times the sum of the other two divisors. Then $n=$
5.95.

Notice that a positive integer with exactly four positive divisors must be of the form $p^{3}$ or $p q$ (where $p$ and $q$ are primes, $p \neq q$).

In the first case, all positive divisors are $1, p, p^{2}, p^{3}$, then $1+p^{3}=4\left(p+p^{2}\right)$, but $p \nmid \left(1+p^{3}\right)$, which is a contradiction.

In the second case, all positive divisors are $1, p, q, p q$, then
$$
\begin{array}{l}
1+p q=4(p+q) \\
\Rightarrow(p-4)(q-4)=15 .
\end{array}
$$

Thus, $q$ is a prime and $q-4$ is a divisor of 15. Hence, $q \in\{5,7,19\}$.
Since $p$ is also a prime, the only solution is $\{p, q\}=\{5,19\}$. Therefore, $n=95$.