Given an equation $$x^4+y^4+z^4+1=4xyz$$Find out the number of possible ordered tuple $(x,y,z)\mid x,y,z\in\Bbb{R}$.

I am getting it as $(1,1,1),(-1,-1,1),(1,-1,-1),(-1,1,-1)$ so $\boxed{4}$

Is there any other tuple which I am missing?

Any help will be appreciable !

  • 1
    $\begingroup$ Looks fine. AM/GM gives that the absolute value of each of $x,y,z$ is 1. $\endgroup$
    – almagest
    Apr 16, 2016 at 14:49
  • $\begingroup$ Wolfram Alpha confirms your answer. $\endgroup$
    – Vepir
    Apr 16, 2016 at 14:54

1 Answer 1


By Am/Gm we have $$\frac{x^4+y^4+z^4+1}{4}\geq xyz$$ . now we know the minima of arithmetic mean and maxima of geometric mean is achieved when numbers are equal or their $mod$ is equal as here $4$th power is used. Thus all positive $1$ or two negative $1$ are permissible hence total answers are $1+{3\choose 2}=1+3=4$

  • $\begingroup$ Nice! I think ur ri8 :-) $\endgroup$
    – ashi
    Apr 16, 2016 at 16:50

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