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.

Guest Jul 8, 2022

#1**+1 **

\(n^2-3n+2 = p\\ n^2-2n-n+2 = p\\ n(n-2)-1(n-2) = p\\ (n-2)(n-1) = p\\ \)

**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.**

sayemsub15 Jul 8, 2022