$\sqrt{x}-\sqrt{z+y}=\sqrt{y}-\sqrt{z+x}=\sqrt{z}-\sqrt{x+y}$ At a recent maths competition one of the questions was to find for which $x,y,z$ this equation holds true:
$$\sqrt{x}-\sqrt{z+y}=\sqrt{y}-\sqrt{z+x}=\sqrt{z}-\sqrt{x+y}$$
where $x,y,z \in \mathbb{R} \cup \{0\}$. So how am I supposed to approach this problem?
Also sorry for not explaining my personal progress on this problem, but there simply is none.
 A: Squaring, simplifying and squaring again gives
$$x(y+z)=y(z+x)=z(x+y)$$
which is $$xy=yz=zx$$ or $$x=y=z.$$

If we assume that $x=0$, then $yz=0$. Then WLOG, $y=0$ and the initial equation says $z=0$.
A: $$\sqrt{x}-\sqrt{z+y}=\sqrt{y}-\sqrt{z+x}=\sqrt{z}-\sqrt{x+y} \Rightarrow$$
$$\sqrt{x}+\sqrt{x+y}=\sqrt{z}+\sqrt{z+y}$$
Let $f(t)=\sqrt t+ \sqrt{t+y},$ where $y -$ fixed. 
$f(t) -$ increasing function $\Rightarrow$ if $f(t_1)=f(t_2)$ then $t_1=t_2$. So $x=z$.
Similarly, $x=y$. So $$x=y=z$$
A: Take square on both sides?
An obvious solution would be $x = y = z$
A: $x \ge 0; y \ge 0; z \ge 0$.
Suppose $z = 0$
Then $\sqrt{x} - \sqrt{y} = \sqrt{y} - \sqrt{x} = - \sqrt{x+y} \implies \sqrt{x} = \sqrt{y} \implies x = y; x + y = 0 \implies x = y = z = 0$.
Likewise if $x = 0$ or $y = 0$ then $x = y = z = 0$ by the same argument so either they all equal 0 or none do.
So assume no $x, y, $ or $z$ = 0:
$\sqrt{x}-\sqrt{z+y}=\sqrt{y}-\sqrt{z+x}=\sqrt{z}-\sqrt{x+y} \implies$
$(\sqrt{x}-\sqrt{z+y})^2=(\sqrt{y}-\sqrt{z+x})^2=(\sqrt{z}-\sqrt{x+y})^2 \implies$
$x+y+z -2(\sqrt{x}\sqrt{z+y})=x+y+z -2(\sqrt{y}\sqrt{z+x})=x+y+z -2(\sqrt{z}\sqrt{x+y}) \implies$
${x}({z+y})={y}({z+x})={z}({x+y}) \implies$
$xz + xy = yz + xy = xz + yz \implies$
$xz = yz; xy = xz; xy = yz $
Thus $x = y; y = z$ and $x = z$.
So any $x = y = z \ge 0$ will work.
