Call a number $n\in\mathbb{N}$ "good" if $d+1|n+1,$ where $d$ is any divisor of $n.$ Find all good numbers.
One immediate thing we see is that $1$ is a good number because the only divisor of $1$ is $1$ and $1+1|1+1$ trivially. We see that $2$ isn't good as $1+1$ doesn't divide $2+1.$ However, any other prime works. The only divisors of a prime $p$ are $1$ and $p.$ Clearly, if $p≠2,$ then $2|p+1.$ Also, $p+1|p+1.$ So, all odd primes and the number $1$ are good.
I think there are no other good numbers. Clearly, any good number must be odd. Else, we would have that $2$ divides an odd number. Let $n\in\mathbb{N}$ be an odd composite number that is good. Take the largest divisor of $n$ that isn't $n$ itself and call it $k.$ Such a $k$ exists precisely because $n$ is composite. From how we've defined $k,$ we have $\sqrt{n}≤k<n.$ Now, $n=km$ for some $m\in\mathbb{N}.$ Now, from $k+1|km+1,$ we get $km+1=t(k+1)$ for some $t\in\mathbb{N}.$ Reduce modulo $k$ to get $t\equiv_k1.$ Hence, $t=kb+1$ for some whole number $b.$ Sub this in to get $km+1=(kb+1)(k+1)=k^2b+kb+k+1.$ Suppose $b≠0.$ Then, $k^2b+kb+k+1≥nb+\sqrt{n}b+\sqrt{n}+1≥n+2\sqrt{n}+1>n+1.$ This gives $n+1>n+1,$ a contradiction. Hence, $b=0.$ This gives $t=1,$ and hence $m=1.$ This gives $n=k,$ contrary to our definition of $k.$ Hence, composite numbers aren't good.
This means that $1$ along with all odd primes are the only good numbers.
Does this look good (pun intended)? Any other ways to go about this?
Alternate Proof:
Let us assume a composite number $n = pk$ where $p$ is smallest prime and $k > 1$.
then
$$k+1|pk+1$$
$$k \equiv -1 \pmod{k+1}$$ We get $p \equiv 1 \pmod{k+1}$
$$k \geq p$$
So, $p-1 = 0$ but $p = 1$ is not a prime.
Therefore, $k \leq 1$.
$k = 1$ gives prime
$k = 0$ is removed as it does not satisfy.