Track / Stage 5 / 322 of 400 #922 of 1964

Problem 922

AIME late

Number theory Difficulty 5.7 Prove it

13.434 Prove that the cube of the largest of three consecutive natural numbers cannot be equal to the sum of the cubes of the other two numbers.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution. Let $n-1, n, n+1$ be arbitrary consecutive natural numbers. Suppose that $(n+1)^{3}=n^{3}+(n-1)^{3} \Rightarrow n^{3}=2\left(3 n^{2}+1\right) \Rightarrow n=2 k$ , i.e., $4 k^{3}=12 k^{2}+1$ . A contradiction is obtained, since $12 k^{2}+1$ is an odd number.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.