Fermat 2006 Q24

Problem Statement

If $a$ and $b$ are positive integers such that $\dfrac{1}{a} + \dfrac{1}{2a} + \dfrac{1}{3a} = \dfrac{1}{b^2 - 2b}$, then the smallest possible value of $a+b$ is

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Solution

$$a, b \in \mathbb{N}$$ $$\dfrac{1}{a} + \dfrac{1}{2a} + \dfrac{1}{3a} = \dfrac{1}{b^2 - 2b}$$ $$1 + \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{a}{b(b-2)}$$ $$\dfrac{11}{6} = \dfrac{a}{b(b-2)} \implies 6 \; \| \; b(b-2)$$ $$min(a + b) \text { is when b is as small as possible} \implies \text { a is as small as possible}$$ $$min(b) = 6 \implies a = 44 \implies min(a + b) = 50$$