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Given $$a^6 + b^6 + c^6 = x^6 + y^6 + z^6$$

prove that $$a^2 + b^2 + c^2 - x^2 - y^2 - z^2 \equiv 0 \bmod{9}$$

I was thinking of using $n^6 \pmod{27}$ and showing both sides have the same pattern but it's getting really confusing..

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  • $\begingroup$ consider $x\equiv 0,1,2,3,4,5,6,7,8 \mod 9$ and compute $x^6\mod 9$ $\endgroup$ – Dr. Sonnhard Graubner Dec 24 '15 at 4:13
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    $\begingroup$ Note that $u^6\equiv 1,0\pmod 9$ for all $u$. So this means that the number of $\{a,b,c\}$ that are divisible by $3$ is the same as the number of $\{x,y,z\}$ that are divisible by $3$. $\endgroup$ – Thomas Andrews Dec 24 '15 at 4:14
  • $\begingroup$ @Dr.SonnhardGraubner : Instead of ^6\mod 9, try ^6\bmod 9. Then instead of $^6\mod 9$ you'll see $x^6\bmod 9$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 24 '15 at 4:15
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    $\begingroup$ It's gonna take more than modular arithmetic, because this theorem is not true of $a^6+b^6+c^6\equiv x^6+y^6+z^6\pmod{9}$. For example, $2^6+0^6+0^6\equiv 1^6+0^6+0^6\pmod{9}$. So you need something about equality. $\endgroup$ – Thomas Andrews Dec 24 '15 at 4:18
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    $\begingroup$ @Shailesh Nothing I wrote is a complete answer. I actually showed why you can't use just my first comment as the answer. $\endgroup$ – Thomas Andrews Dec 24 '15 at 4:20
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$(9n+a)^6=a^6\bmod27$,
so only worry about numbers between $-4$ and $4$.
Their sixth powers are $19,0,10,1,0,1,10,0,19\pmod{27}$
and their squares are $7,0,4,1,0,1,4,7\pmod{9}$
The sixth powers are either $0$ or $9A+1$; the squares are either $0$ or $3A+1$, the same $A$.

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Write $$ a^6+b^6+c^6-x^6-y^6-z^6 = (a^2-x^2)(a^4+a^2x^2+x^4) + (b^2-y^2)(b^4+b^2y^2+y^4) + (c^2-z^2)(c^4+c^2z^2+z^4) ; $$ by Thomas's comment, we can also pair off in such a way that $a,x$ are either both $\equiv 0\bmod 3$ or both $\not\equiv 0\bmod 3$, and similarly for $b,y$ and $c,z$.

The non-zero squares $\bmod 9$ are $1,4,7$, and now some experimentation shows that $a^4+a^2x^2+x^4\equiv 3\bmod 9$ if $a,x\not\equiv 0\bmod 3$. Thus the expression from above becomes $$ 3(a^2-x^2+b^2-y^2+c^2-z^2) $$ when considered $\bmod 9$; note that this is also correct when the other case ($a,x\equiv 0\bmod 3$) applies. Since the difference of the sums of squares is $\equiv 0\bmod 3$ (by Thomas's comment again, and since $u^2\equiv 1\bmod 3$ for $u\not\equiv 0\bmod 3$), the claim follows now.

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