4
$\begingroup$

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?

$\endgroup$
5
  • 1
    $\begingroup$ Doesn't $x=y=z$ Imply $2x = \sqrt{x^{2} + 2018}$ from which you can solve for $x$? $\endgroup$ Mar 30 '18 at 16:16
  • 1
    $\begingroup$ 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 $f(x) = f(y) = f(z) \implies \cdots$ $\endgroup$ Mar 30 '18 at 16:22
  • 1
    $\begingroup$ Indeed, we get $$x=y=z=\sqrt{\frac{2018}{3}}$$ $\endgroup$ Mar 30 '18 at 16:25
  • $\begingroup$ @PrathyushPoduval yes, thanks for that, i was meant that way. I have edited the post $\endgroup$
    – Arief
    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 $z \le (m-1)/2$. $\endgroup$ Mar 30 '18 at 16:29
1
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ 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
    Mar 30 '18 at 16:42
  • $\begingroup$ Silly me. You are correct. $\endgroup$
    – Bill Cook
    Mar 30 '18 at 17:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.