AoPS - Problem

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Tags: number theory, contests, Tricubic numbers

ZETA_in_olympiad

24.04.2022 18:56

We say that a positive integer $k$ is tricubic if there are three positive integers $a, b, c,$ not necessarily different, such that $k=a^3+b^3+c^3.$ a) Prove that there are infinitely many positive integers $n$ that satisfy the following condition: exactly one of the three numbers $n, n+2$ and $n+28$ is tricubic. b) Prove that there are infinitely many positive integers $n$ that satisfy the following condition: exactly two of the three numbers $n, n+2$ and $n+28$ are tricubic. c) Prove that there are infinitely many positive integers $n$ that satisfy the following condition: the three numbers $n, n+2$ and $n+28$ are tricubic.

Note that perfect cubes can only be $-1, 0, 1 \pmod{9}$, so all numbers that are $4$ or $5 \pmod{9}$ cannot be tricubic. (a) Let $n = (3k+1)^3 + (3k+1)^3 + (3k+1)^3$ for a positive integer $k$. Then $n + 2 \equiv 5 \pmod{9}, n + 28 \equiv 4 \pmod{9}$. So only $n$ is tricubic. (b) Let $n = (3k+1)^3 + 10^3 + 18^3$ for a positive integer $k$. Since $1^3 + 19^3 - 10^3 - 18^3 = 28$, $n + 28 = (3k+1)^3 + 1^3 + 19^3$. Note that $n + 2 \equiv 4 \pmod{9}$, so only $n$ and $n + 28$ are tricubic. (c) Let $n = (3k)^3 + (4k)^3 + (5k)^3 = (6k)^3$ for a positive integer $k$. Then $n + 2 = (6k)^3 + 1^3 + 1^3$ and $n + 28 = (6k)^3 + 1^3 + 3^3$. So all three numbers are tricubic.