how would I prove that if 9 divides $x^3+y^3+z^3$ then $3$ divides $xyz$?
I've thought about it and don't really know where to start. So far, I know that $x^3+y^3+z^3$ is congruent to $0 (\mod 9)$.
I don't know where to go from here.
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WHY does it only work for consecutive integers? The answer to that may supply a clue to the proof. – Mathemagician1234 Mar 04 '15 at 07:43
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It need not only work for consecutive integers. For example take $x=3, y=6, z=12$. – Anurag A Mar 04 '15 at 07:45
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@AnuragA Thank you,that's why it sounded fishy to me without a proof of that statement. – Mathemagician1234 Mar 04 '15 at 07:47
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Observe that for any integer $a$ we have $a^3 \equiv 0, \pm 1 \pmod{9}$. So for $x^3+y^3+z^3 \equiv 0 \pmod{9}$ we can only have the following scenarios:
- Each of $x^3, y^3$ and $z^3$ are $0 \bmod{9}$.
- One of them is $0 \bmod{9}$ and the remaining two are $1 \bmod{9}$ and $-1 \bmod{9}$.
But in both cases we have at least one (say) $x^3 \equiv 0 \pmod{9}$. This implies $x^3 \equiv 0 \pmod{3}$. Consequently $x \equiv 0 \pmod{3}$. This implies $xyz \equiv 0 \pmod{3}$.
Anurag A
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How would you prove that $a^3 \equiv 0, \pm 1 \pmod{9}$ for every n?Without proving this first, you're whole approach is suspect. I'd assume induction. – Mathemagician1234 Mar 04 '15 at 07:47
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No need of induction. Any integer when divided by $9$ will leave a remainder from the set ${0,1,2, \ldots ,8}$. Thus when we compute $x^3 \bmod{9}$ the residues will be among ${0^3,1^3,2^3, \ldots ,8^3}$. Now modulo $9$ this set has only three distinct elements ${0,1,-1}$. – Anurag A Mar 04 '15 at 07:49
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Very nice common sense proof! I gotta review my number theory. Melvyn Nathanson would have made a clown out of me for asking that question in class.......lol – Mathemagician1234 Mar 04 '15 at 07:51