Three consecutive positive integers k, l and m are such that l2 = 3(k+m). Find the value of m

Three consecutive positive integers k, l and m are such that l2 = 3(k+m). Find the value of m

Answer Details

Let's start by translating the given problem into equations. First, we know that the three integers are consecutive, so we can express them as k, k+1, and k+2. Second, we know that l^2 = 3(k+m). We can substitute k+1 for l to get (k+1)^2 = 3(k+m). Expanding (k+1)^2, we get k^2 + 2k + 1 = 3(k+m). Simplifying, we get k^2 - k(6-3m) + 3m - 1 = 0. We can use the quadratic formula to solve for k: k = [(6-3m) ± sqrt((6-3m)^2 - 4(3m-1))]/2. Simplifying under the square root, we get sqrt(9m^2 - 36m + 37). Since k is a positive integer, the expression under the square root must be a perfect square. We can check the perfect squares between 1 and 37 and find that only 4, 16, and 25 are possibilities. Solving for m in each case gives us three potential solutions: 4, 5, and 7. However, only one of these solutions satisfies the original equation l^2 = 3(k+m), and that is m=4. Therefore, the answer to the problem is m=4.