<s>[INST] $\sqrt{x}-\sqrt{z+y}=\sqrt{y}-\sqrt{z+x}=\sqrt{z}-\sqrt{x+y}$ 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.
 [/INST] $x \ge 0; y \ge 0; z \ge 0$.
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.
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