Romanian Mathematical Olympiad

Since $p$ is a prime divisor of the number of automorphisms of $G$, some automorphism $f$ has order $p$. Since $f$ is a permutation of the set $G \setminus \{e\}$, it follows that $f$ is a cycle of length $p$, so $G \setminus \{e\} = \{x, f(x), \dots, f^{p-1}(x)\}$, whatever $x$ in $G \setminus \{e\}$. On the other hand, $|G| = p+1$ is even, so $G$ contains an element $x_0$ of order $2$. Hence $\text{ord} f^k(x_0) = 2$, $k = 0, \dots, p-1$, so $x^2 = e$ for all $x$ in $G$, and $p+1 = 2^n$ for some integer $n \ge 2$. Consequently, $p = 2^n - 1 \equiv 3 \pmod 4$.