This is problem 19 from AMC 8, 2007.
Pick two consecutive integers whose sum is less than 100. Square both of those integers and then find the difference of the squares. Which of the following could be the difference? 2, 64, 79, 96, 131.
Solution
The key to the solution is the well-known formula for the differences of two squares
a² - b² = (a - b)(a + b).
Let a be the larger of the two: a > b, implying a = b + 1, since the two are consecutive. Equivalently, a - b = 1. It follows that the difference of the squares equals the sum of the two numbers. Since the sum is required to be less than 100 it could not be 131,2,64,79,96,131.
Of the two consecutive numbers one is odd and one is even, implying that the sum of the two is odd. Among the four remaining possibilities there is only one odd number, 79,2,64,79,96,131, which is necessarily is the answer to the problem.
We may actually find the two numbers, although this is not required to solve the problem. The smallest of the two, b, is found from 2b + 1 = 79,2,64,79,96,131, giving b = 39 and then a = 40.
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