Then \(n\) is called a prime number if \(n\) has exactly two positive divisors, \(1\) and \(n.\) Then \( q \) is also a prime divisor of \(n\) and \( q < m < \sqrt{n} < p.\) This is a contradiction. B...Then \(n\) is called a prime number if \(n\) has exactly two positive divisors, \(1\) and \(n.\) Then \( q \) is also a prime divisor of \(n\) and \( q < m < \sqrt{n} < p.\) This is a contradiction. But this is impossible since there is no prime that divides 1 and as a result \(q\) is not one of the primes listed. Consider the sequence of integers \[(n+1)!+2, (n+1)!+3,...,(n+1)!+n, (n+1)!+(n+1)\]