Prove $x+y+z$ divides $x^2+y^2+z^2$ where x,y,z are consecutive terms in a geometric series The natural numbers $x,y,z$ are consecutive terms in a geometric series. Proves that $x+y+z$ divides $x^2+y^2+z^2$.
$x = ar\\
y = ar^2 = xr\\
z = ar^3 = xr^2$
So..
$\begin{align}
x + y + z & = x + xr + xr^2 \\
 & = x(1+r+r^2)\end{align}$
$\begin{align}
x^2 + y^2 + z^2 & = x^2 + x^2r^2 + x^2r^4\\
& = x^2(1+r^2+r^4)\end{align}$
I can see that x is a factor for both but I've no idea where to go from here

An alternate solution is to observe that 
$$x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz)$$ 
and
$$xy+xz+yz=x^2r(1+r+r^2)$$
