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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.

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4 Answers 4

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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 $z=0$.

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  • $\begingroup$ But $(x,y,z)=(1,0,0)$ also satisfies $xy=yz=zx$.... $\endgroup$ Mar 18, 2016 at 15:15
  • $\begingroup$ @BarryCipra: I wasn't done :) $\endgroup$
    – user65203
    Mar 18, 2016 at 15:17
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$$\sqrt{x}-\sqrt{z+y}=\sqrt{y}-\sqrt{z+x}=\sqrt{z}-\sqrt{x+y} \Rightarrow$$ $$\sqrt{x}+\sqrt{x+y}=\sqrt{z}+\sqrt{z+y}$$ 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 $$x=y=z$$

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Take square on both sides? An obvious solution would be $x = y = z$

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$x \ge 0; y \ge 0; z \ge 0$.

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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