AIME 1985 P3

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Problem Statement

Find $c$ if $a$, $b$, and $c$ are positive integers which satisfy $c=(a + bi)^3 - 107i$, where $i^2 = -1$.

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Solution

$$c = (a^2 + 2abi + b^2i^2)(a + bi) - 107i$$ $$c = a^3 + 2a^2bi + ab^2i^2 + a^2bi + 2ab^2i^2 + b^3i^3 - 107i$$ $$c = a^3 + 3a^2bi + 3ab^2i^2 + b^3i^3 - 107i$$ $$c = (a^3 - 3ab^2) + (3a^2bi - b^3i - 107i)$$ $$\text{Since } c \in \mathbb{Z} \implies 3a^2bi - b^3i - 107i = 0$$ $$3a^2b - b^3 = 107$$ $$b(3a^2 - b^2) = 107$$ $$\text{Since 107 is a prime number }\implies b = 1, 3a^2 - b^2$$ $$\implies 3a^2 - (1^2) = 107 \implies a = 6$$ $$\therefore c = (a^3 - 3ab^2) + (3a^2bi - b^3i - 107i) = 6^3 - 3(6)(1^2) + 0 = 198$$