Croatian Mathematical Olympiad

Since $n$ divides $p-1$, there exists a positive integer $a$ such that $p-1 = an$. We also have $p-1 \ge n$.

From the condition that $n^3-1 = (n-1)(n^2+n+1)$ is divisible by the prime number $p$, it follows that $n^2+n+1$ is divisible by $p$. Indeed, $1 \le n-1 < n+1 \le p$, so $n-1$ cannot be divisible by $p$.

Hence $an+1 \mid n^2+n+1$. This implies that $1 \le a \le n+1$ (because if $a \ge n+2$, then $an+1 \ge (n+2) \cdot n+1 = n^2+2n+1 > n^2+n+1$, which is impossible).

From the same divisibility it follows that $an+1 \mid a \cdot (n^2+n+1) - n \cdot (an+1) = (a-1)n+a$, which is positive, so we must have $(a-1)n+a \ge an+1$, i.e. $a \ge n+1$.

It follows that $a = n+1$ and $p = n^2+n+1$.

Hence $4p-3 = 4n^2+4n+1 = (2n+1)^2$, which finishes the proof.