<s>[INST] 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
 [/INST] WLOG let $x=\frac 1r, y=1, z=r$. 
Then  $x+y+z=r+1+\frac 1r$.
Consider 
$$\left(\frac 1r+r+1\right)\left(\frac 1r+r-1\right)=\left(\frac 1r+r\right)^2-1^2\\
=\left(\frac 1{r^2}+2+r^2\right)-1\\
=\frac 1{r^2}+1+r^2\\
\Rightarrow \quad (x+y+z)(x-y+z)=x^2+y^2+z^2$$
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