Let $a, b,c \in \mathbb{Z}: 9 \mid a^3 + b^3 + c^3$
I need to show that $3 \mid a \text{ or } 3 \mid b \text{ or } 3 \mid c$
I don´t have any clue how should I approach this problem. Could someone give my hints or name some theorems that should I use.
Thank you in advance
Hint:
Write the list of cubes in $\mathbf Z/9\mathbf Z$. It will be faster to write the elements of $\mathbf Z/9\mathbf Z$ as $$\{0,,\pm1,\pm 2,\pm 3,\pm4\}.$$
If $x\not\equiv 0\pmod{3}$, then $x^3\equiv \pm 1\pmod{9}$, since
$ (3x\pm 1)^3 = 9X\pm 1$.
The sum of three numbers in $\{-1,+1\}$ cannot be $\equiv 0\pmod{9}$, hence $a^3+b^3+c^3\equiv 0\pmod{9}$ implies that at least one number among $a,b,c$ is a multiple of three.
Since $\phi(9)=6$, (and $9$ is a prime square) we know that, $a^3 \equiv (k\in\{-1,0,1\}) \bmod 9$ (and similarly for $b$ and $c$). Therefore we must have at least one of $a^3, b^3, c^3 \equiv 0 \bmod 9$ , say $d^3$, and since $3$ is prime, $3^2 \mid d^3 \implies 3 \mid d$.