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..
Solution 1:
$(9n+a)^6=a^6\bmod27$,
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 $A$.