3. Find all natural numbers $n$ with the following property: for every positive divisor $d$ of $n$, the number $d+1$ is a divisor of $n+1$.
Solution. If $n$ is an odd prime number or one, the divisors of $n$ are 1 and $n$, but then $n+1$ will be an even number for which $1+1=2$ and $n+1$ are certainly its divisors.

If $n$ is an even number, then $n+1$ will be an odd number and $1+1=2$ will not be a divisor of $n+1$, while one is a divisor of $n$, so $n$ cannot be an even number.

If $n$ is an odd composite number, then

$$
\begin{array}{ll}
n=k l \Rightarrow n+1=k l+1=(k+1)(l-s)=(l+1)(k-t) & \Rightarrow \\
k l+1=k l+l-k s-s=k l+k-l t-t & \Rightarrow \\
1=l-s(k+1)=k-t(l+1) & \Rightarrow \\
l>k+1, k>l+1 & \Rightarrow \\
l>k+1>l+2 &
\end{array}
$$

which is not possible. Therefore, the required numbers are 1 and all odd prime numbers.