Find all pairs of $(x,y,z)$, of real numbers, such that
$$ x + y = \sqrt{z^{2} + 2018} $$ $$ x + z = \sqrt{y^{2} + 2018} $$ $$ y + z = \sqrt{x^{2} + 2018} $$
An attempt : Squaring we get $$ (x + y)^{2} = z^{2} + 2018 \implies (x + y)^{2} - z^{2} = 2018 $$ $$ (x + z)^{2} = y^{2} + 2018 \implies (x + z)^{2} - y^{2} = 2018 $$ $$ (y + z)^{2} = x^{2} + 2018 \implies (z + y)^{2} - x^{2} = 2018 $$ which also means $$ (x + y - z)(x + y + z) = 2018 $$ $$ (x + z - y)(x + y + z) = 2018 $$ $$ (z + y - x)(x + y + z) = 2018 $$ so $$ \frac{2018}{x+y-z} = \frac{2018}{x+z-y} = \frac{2018}{y+z-x} $$
$$(x+y-z) = (x+z-y) = (y+z-x)$$ $$y-z = z-y \implies z = y $$ $$x-y= y-x \implies x=y $$ so my answer is $$(x, y, z), \:\:\: x=y=z $$ but with the 3 initial equations, we must also have $$ (x+y) = \sqrt{z^{2} + 2018} \implies 4x^{2} = x^{2} + 2018 $$ or $$ x^{2} = 2018/3 $$ so the solution is $$(x, y, z), \:\:\: x=y=z = \sqrt{2018/3} $$
Is this sufficient already? are there better techniques?
At the stage you have... $$ (x + y - z)(x + y + z) = 2018 $$ $$ (x + z - y)(x + y + z) = 2018 $$ $$ (z + y - x)(x + y + z) = 2018 $$ This implies... $$x+y-z=x+z-y=z+y-x=\dfrac{2018}{x+y+z}$$ ...unless $x+y+z=0$.
Notice that $x+y+z=0$ implies $x+y=-z$ and so $-z=\sqrt{z^2+2018}$ which means $z^2=z^2+2018$ and so $z=0$. Likewise, $x=y=0$ (contradiction since $(0,0,0)$ isn't a solution).
Thus $x+y-z=x+z-y=z+y-x$ and so you got that $x=y=z$. But then $x+y=\sqrt{z^2+2018}$ implies that $2z=\sqrt{z^2+2018}$ and so $4z^2=z^2+2018$ and so $x=y=z = \pm \sqrt{\dfrac{2018}{3}}$. But $x+y $ etc. are square roots (thus non negative). So there is only one solution.