RMO 2004 Questions, Solutions, Discussions

Note that \(p_1p_2 \dots p_n + 1\) is odd for any \(n\geq 1\), and hence \(p_n\) is odd for any \(n\geq 2\). Since \(p_1 = 2\) and \(p_2 = 3\), it follows that for any \(n\geq 2\), the integer \(p_1p_2 \dots p_n + 1\) is not divisible by any one of \(2\) and \(3\). So the least prime divisor of \(p_1p_2 \dots p_n + 1\) is at least \(5\) for any \(n\geq 2\). If possible, suppose \(5\) is the largest prime divisor of \(p_1p_2 \dots p_n + 1\) for some integer \(n\geq 2\). This yields \[ p_1p_2 \dots p_n + 1 = 5^k \] for some \(k\geq 1\). This implies that \(4\) divides \(p_1p_2 \cdots p_n\), which is impossible since \(p_1 = 2\), and \(p_r\) is odd for any integer \(r \geq 2\). This shows that \(p_{n+1}\) is not equal to \(5\) for any integer \(n\geq 2\). Consequently, it follows that \(p_n\neq 5\) for any integer \(n\geq 1\).