Number Theory Challenge – 12/12/2025

Proof: Let $a=3A+x$, $b=3B+y$, $c=3C+z$, where $A$, $B$, $C$ are integers, $a\pmod 3=x$, $b\pmod 3=y$, $c\pmod 3=z$. Therefore $0\le x,y,z\le 2$.

Without loss of generality, we assume that $0\le x\le y\le z\le 2$. We have

$$(a^3+b^3+c^3)\pmod 9=((3A+x)^3+(3B+y)^3+(3C+z)^3)\pmod 9$$ $$\ \ \ =(x^3+y^3+z^3)\pmod 9$$

If $9|(a^3+b^3+c^3)$, then $9|(x^3+y^3+z^3)$. If $x\ne 0$, then none of $x$,$y$,$z$ is $0$, and none of $a$,$b$,$c$ is divisible by $3$, the $(x,y,z)$ values can only be $(1,1,1)$, $(1,1,2)$, $(1,2,2)$, $(2,2,2)$, which result in $(x^3+y^3+z^3)$ values as $3$, $10$, $17$, $24$, with none of them divisible by $9$. Therefore, $x=0$, and at least one of $a$, $b$, $c$ is divisible by $3$.