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Number theory Difficulty 4.4 AIME Prove it Hong Kong

Show that the product of three consecutive integers is the sum of four integer cubes.

Solution

Note that the product of three consecutive integers is divisible by $6$ . Let the product be $6k$ for some $k \in \mathbb{Z}$ . Then we easily check that
$(k + 1)^3 + (k - 1)^3 + (-k)^3 + (-k)^3 = 6k.$
This completes the proof.

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