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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?

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    $\begingroup$ Doesn't $x=y=z$ Imply $2x = \sqrt{x^{2} + 2018}$ from which you can solve for $x$? $\endgroup$ – Jim Haddocc Mar 30 '18 at 16:16
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    $\begingroup$ The equation can be rewritten as $x + y + z = f(x) = f(y) = f(z)$ where $f(t) = t + \sqrt{t^2+2018}$. The function $f$ is strictly increasing, so $f(x) = f(y) = f(z) \implies \cdots$ $\endgroup$ – achille hui Mar 30 '18 at 16:22
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    $\begingroup$ Indeed, we get $$x=y=z=\sqrt{\frac{2018}{3}}$$ $\endgroup$ – Dr. Sonnhard Graubner Mar 30 '18 at 16:25
  • $\begingroup$ @PrathyushPoduval yes, thanks for that, i was meant that way. I have edited the post $\endgroup$ – Arief Anbiya Mar 30 '18 at 16:26
  • $\begingroup$ Just the condition that $z^2+m$ is a square implies that$z^2+m \ge (z+1)^2$ or $z \le (m-1)/2$. $\endgroup$ – marty cohen Mar 30 '18 at 16:29
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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.

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    $\begingroup$ Thanks for the answer, but i think negative value of $x=y=z < 0$ is not allowed. This is because the 3 initial equations hold. $\endgroup$ – Arief Anbiya Mar 30 '18 at 16:42
  • $\begingroup$ Silly me. You are correct. $\endgroup$ – Bill Cook Mar 30 '18 at 17:15

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