# $3\mid a^3+b^3+c^3$ if only if $3\mid a+b+c$

Prove the following equivalence: $3\mid (a^3 + b^3 + c^3)$ if and only if $3\mid (a + b + c)$.

My try:

I know $a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) + 3abc$, but I can't seem to proceed from here.

Thanks all!

• Try to show that $a^2+b^2+c^2–ab–bc–ca$ is not divisible by 3. Or follow the answers posted here. Mar 21, 2017 at 11:40

Fact is that, $n^3−n=(n−1)n(n+1)$ being a product of three consecutive integers is a multiple of $3$.

Hence $3$ divides $(a^3−a)+(b^3−b)+(c^3−c)$ for any triple of integers $a,b,c$. Rearranging the terms, we have that $3$ dividing $(a^3+b^3+c^3)−(a+b+c)$.

• Thanks for an alternative solution. Mar 21, 2017 at 11:43

Very short with Lil' Fermat:

for any $x\in\mathbf Z$, $\;x^3\equiv x\mod 3$. Hence $\;a^3+b^3+c^3\equiv a+b+c\mod 3$.

Compare $a^2+b^2+c^2-ab-bc-ca$ with $(a+b+c)^2$

By Fermat's Little Theorem we know that $3$ divides $x^3-x$ for any $x \in \mathbb{Z}$

Which means $(a^3-a)+(b^3-b)+(c^3-c) = (a^3+b^3+c^3)-(a+b+c)$ is divisible by $3$.

The conclusion is easy to derive now.

• $1^3 = 1 \mod 3$
• $2^3 = 2 \mod 3$
• $3^3 = 3 \mod 3$

This means $x = x^3 \mod 3$ for any integer $x$. This also means that if $a^3 + b^3 + c^3 = 0 \mod 3$ then $a + b + c = 0 \mod 3$.

Thus it is divisible by three.