# are these the only answers of $x^4+y^4+z^4+1=4xyz$?

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 !

• Looks fine. AM/GM gives that the absolute value of each of $x,y,z$ is 1. Apr 16, 2016 at 14:49
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 $1+{3\choose 2}=1+3=4$