Stars of Mathematics Competition

We prove that the numbers that have the given property are $1$ and the odd prime numbers. It is clear that all these numbers do indeed have the desired property and also that $2$ does not have it.

Conversely, let us consider a composite number $n$ and prove that it does not have the given property. If $n$ is composite, then $n=ab$ such that $1<a\le b<n$. It follows that $b+1$ divides $n+1$, i.e. there exists $c\in \mathbb{Z}$ such that $c(b+1)=n+1=ab+1$. We obtain that $b$ divides $c-1$. Obviously, $c>1$. We deduce that $c-1\ge b$, i.e. $c\ge b+1$. Then $ab+1=c(b+1)\ge (b+1{)}^2$, which means that $ab+1\ge b^2+2b+1$, leading to $a\ge b+2$, which contradicts $a\le b$.

In conclusion, no composite number does satisfy the requirements.