Library / Hong Kong competition problems /1 of 42

Hong Kong competition problems, 2008

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.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.