$6$ divides $a+b+c$ if and only if $6$ divides $a^{3}+b^{3}+c^{3}$

Question

Prove that $6$ divides $a+b+c$ if and only if $6$ divides $a^{3}+b^{3}+c^{3}$

Here's my attempt:

As $6$ divides $a+b+c$ , $a+b+c$ must be even. This implies, $a^{3}+b^{3}+c^{3}$ is also even. Also, By Fermat's theorem $$a^3 \equiv a\pmod 3$$ Similarly $$b^3 \equiv b\pmod 3$$ and $$c^3 \equiv c\pmod 3$$. Therefore, $$a^{3}+b^{3}+c^{3} \equiv {a+b+c} \pmod 3$$ This implies that, $3$ divides $a^{3}+b^{3}+c^{3}$ and $a^{3}+b^{3}+c^{3}$ is even. Therefore , $6$ divides $a^{3}+b^{3}+c^{3}$. Does this look correct? Also, Could you please help me write the proof for converse part. Thank you.

Answer 1

Do both directions of the proof together.

For any integer $x$, $x^3 \equiv x\pmod 3$ and $x^3 \equiv x\pmod 2$. Therefore $x^3 \equiv x\pmod 6$.

Then $a^{3}+b^{3}+c^{3}\equiv(a+b+c)\pmod 6$.

So $6$ divides $a+b+c$ if and only if it divides $a^{3}+b^{3}+c^{3}$.