<s>[INST] $x^7+y^7+z^7$ is divisible by $7^3$, then $x+y+z$ is divisible by $7^2$ 
Let $x, y, z$ be positive integers, and $7 \nmid xyz$. If $7^3|x^7+y^7+z^7$, show that $7^2|x+y+z$.

by Fermat's little theorem, $x^7 \equiv x \pmod7$, then $x^7+y^7+z^7\equiv x+y+z \equiv 0 $ (mod 7)
so we have $7 | (x+y+z)$.
what should I do next?
 [/INST] HINT:
If $x+y\equiv0\pmod7,$ we are done
Otherwise
We can write $z=7a-x-y$
$$x^7+y^7+z^7=x^7+y^7+(7a-x-y)^7$$
$$(7a-x-y)^7\equiv-(x+y)^7+7(7a)(x+y)^6\pmod{7^3}$$
$$\implies-(x+y)^7+7(7a)(x+y)^6\equiv0\pmod{p^3}$$
As $7\nmid(x+y),$  $$x+y\equiv-49a\pmod{7^3}$$
which is impossible as $7\nmid(x+y)$
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