Problem 18 - Entrance Test

We want to find the number of positive integers n such that n^2 + 1 is divisible by n + 1. This means (n^2 + 1) / (n + 1) must be an integer. We can use polynomial long division or algebraic manipulation to simplify the expression n^2 + 1 in terms of n + 1. Consider the identity a^2 - b^2 = (a-b)(a+b). We can write n^2 + 1 as n^2 - 1 + 2. The term n^2 - 1 is divisible by n+1: n^2 - 1 = (n-1)(n+1). So, n^2 + 1 = (n-1)(n+1) + 2. For n^2 + 1 to be divisible by n + 1, it must be that (n-1)(n+1) + 2 is divisible by n + 1. Since (n-1)(n+1) is clearly divisible by n + 1, for the entire expression to be divisible by n + 1, the remainder, which is 2, must be divisible by n + 1. Therefore, n + 1 must be a divisor of 2. The divisors of 2 are {1, 2, -1, -2}. Since n is a positive integer, n must be at least 1. This implies that n + 1 must be at least 1 + 1 = 2. So, we only consider the positive divisors of 2 that are greater than or equal to 2: 1. n + 1 = 1 => n = 0. This is not a positive integer. 2. n + 1 = 2 => n = 1. This is a positive integer. Thus, there is only one positive integer n (which is n=1) for which n^2 + 1 is divisible by n + 1. Let's check for n=1: (1^2 + 1) / (1 + 1) = 2 / 2 = 1, which is an integer.