AIME 1999 P3

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Problem Statement

Find the sum of all positive integers $n$ for which $n^2-19n+99$ is a perfect square.

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Solution

$$\text {let } k \in \mathbb{Z} \text { such that } n^2 - 19n + 99 = k^2, n \in \mathbb{N}$$ $$n^2 - 19n + (99 - k^2) = 0$$ $$n = \dfrac{-(-19) \pm \sqrt{(-19)^2 - 4(1)(99 - k^2)}}{2(1)}$$ $$n = \dfrac{19 \pm \sqrt{4k^2 - 35}}{2}$$ $$\text {Since } n \in \mathbb{N} \implies 4k^2 - 35 = m^2, m \in \mathbb{Z}$$ $$4k^2 - m^2 = 35$$ $$(2k - m)(2k + m) = 35 = (1)(35) = (5)(7)$$ $$\text {let } 2k - m = 1, 2k + m = 35 \implies 4k = 36 \implies k = 9 \implies m = 17$$ $$\therefore n = \dfrac{19 + \sqrt{17^2}}{2} = 18 \text { OR } n = \dfrac{19 - \sqrt{17^2}}{2} = 1$$ $$\text {let } 2k - m = 5, 2k + m = 7 \implies 4k = 12 \implies k = 3 \implies m = 1$$ $$\therefore n = \dfrac{19 + \sqrt{1^2}}{2} = 10 \text { OR } n = \dfrac{19 - \sqrt{1^2}}{2} = 9$$ $$\therefore n \in \{1, 9, 10, 18\}$$ $$\therefore \text{ the sum of all possible } n = 1 + 9 + 10 + 18 = 38$$