Find the product of all positive integral values of n such that n^2 - 3n + 2 = p for some prime number p. Note that there is at least one such n.
+2 \(n^2-3n+2 = p\\ n^2-2n-n+2 = p\\ n(n-2)-1(n-2) = p\\ (n-2)(n-1) = p\\ \)
p is a prime equal to the product of two consecutive numbers. Since the product of two consecutive numbers is always even, p is an even prime number so p = 2.So n = 3.
