Find three consecutive odd integers such that the sum of the squares of the first two is 15 less than the square of the third.
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Originally Posted by andrewpark2 Find three consecutive odd integers such that the sum of the squares of the first two is 15 less than the square of the third. let n be the first odd integer, then n + 2 is the next, and n + 4 is the last, we want, $\displaystyle n^2 + (n + 2)^2 = (n + 4)^2 - 15$ now solve for n and you can find the integers
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