$x^7+y^7+z^7$ 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?

Since
$$x^7+y^7+x^7=\sum_{cyc}(x^7-x)+x+y+z,$$
we see that $x+y+z$ is divided by $7$.
In another hand,
$$x^7+y^7+z^7=(x+y+z)^7-7(xy+xz+yz)(x+y+z)^5+7xyz(x+y+z)^4+$$
$$+14(xy+xz+yz)^2(x+y+z)^3-21xyz(xy+xz+yz)(x+y+z)^2-$$
$$-7(xy+xz+yz)^3(x+y+z)+7x^2y^2z^2(x+y+z)+7xyz(xy+xz+yz)^2,$$
which says (see the last term) that $(xy+xz+yz)^2$ is divided by $7$ or
$xy+xz+yz$ is divided by $7$.
Thus, $7x^2y^2z^2(x+y+z)$ is divided by $7^3$ and we are done!
