If two integer triples have the same sum of 6th powers, then their sums of squares agree $\bmod 9$ 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..
 A: $(9n+a)^6=a^6\bmod27$,
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 $A$.
A: 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.
