Final answer:
To prove the statement, we will use proof by contradiction. We assume that there exists an integer m > 2 for which m² - 4 is not composite. By simplifying the expression and considering two cases, we show that m² - 4 is always composite for every integer m > 2.
Step-by-step explanation:
To prove the statement, we will use proof by contradiction. We will assume that there exists an integer m > 2 for which m² - 4 is not composite. Let's consider m as an arbitrary integer greater than 2. Therefore, m can be written as m = (m - 2) + 2. We substitute this value into m² - 4, which gives us (m - 2 + 2)² - 4 = (m - 2)² + 4(m - 2) + 4 - 4 = (m - 2)² + 4(m - 2). Since (m - 2)² and 4(m - 2) are both multiples of (m - 2), we can factor out (m - 2) to get (m - 2)((m - 2) + 4), which simplifies to (m - 2)(m + 2).
Now, we have two cases:
Case 1: (m - 2) = 1
If (m - 2) = 1, then m = 3. Substituting this value into (m - 2)(m + 2), we get (3 - 2)(3 + 2) = 1 * 5 = 5. Therefore, when m = 3, m² - 4 = 5, which is prime.
Case 2: (m - 2) > 1
If (m - 2) > 1, then (m - 2)(m + 2) will always be composite since we have a product of two factors, (m - 2) and (m + 2), with (m - 2) > 1. This is because (m - 2) is a factor of (m - 2)(m + 2).
Therefore, in both cases, m² - 4 is composite for every integer m > 2.