Problem 9 - Entrance Test

The equation given can be factored into (a + b)(a - b) = 15. Since a and b are positive integers, the factors of 15 that make sense for (a + b) and (a - b) would be 15 and 1 or 5 and 3. Considering a > b, the only pair that works for (a + b) and (a - b) where both are positive and their difference makes sense is 5 and 3 (because if a and b were 8 and 7, their difference would be 1 and their sum would be 15, but this doesn't fit our factors directly). This means a + b could be 8 and a - b could be 3 (since 8 * 3 = 24 does not equal 15, this explanation misaligns with factoring 15 directly into the equation). Correctly, if (a+b)(a-b) = 15, we look for factors of 15: 1*15 and 3*5. Given a>b, for the difference and sum to both be positive, we consider pairs that could fit (a+b) and (a-b). For 3*5, if a-b = 3 and a+b = 5, solving these simultaneous equations gives a = 4, b = 1, and indeed a^2 - b^2 = 16 - 1 = 15, and a + b = 4 + 1 = 5. This option is not listed; re-evaluation shows the confusion in factor application and proper equation handling.