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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!

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

5 Answers 5

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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 $(a^3+b^3+c^3)−(a+b+c)$.

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    $\begingroup$ Thanks for an alternative solution. $\endgroup$ Mar 21, 2017 at 11:43
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Very short with Lil' Fermat:

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

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

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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.

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

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.

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