Find the number of triples $(x,y,z)$ of real numbers satisfying the equation $x^4+y^4+z^4+1=4xyz$ QUESTION
Find the number of triples $$(x,y,z)$$ of real numbers satisfying the equation $$x^4+y^4+z^4+1=4xyz$$
I have tried to solve this problem but can't quite understand how to manipulate the given data to reach a clear result. Could someone please explain how I should approach and solve this problem.Thanks :)
 A: The left side is equal to $ (x^{2} − y^{2})^{2}  + 2x^{2}y^{2} + z^{4} − 4xyz $. The presence of $2x^{2}y^{2} $ and $ −4xyz $ suggests the possibility of adding a $ 2z^{2} $ and then completing one more square. So the equation can be rewritten as $ (x^{2} − y^{2})^{2} + (z^{2} − 1)^{2} + 2(xy − z)^{2} = 0 $. This equality can hold only if all three squares are equal to zero. From $ z^{2} − 1 = 0 $ we have $ z = ±1 $, and after a quick analysis we conclude that the solutions are $ (1, 1, 1), (−1, 1, −1), (1, −1, −1), (-1,-1,1) $. Hope it helps.
A: We have, by the AM-GM inequality, that
$$
\frac{1+x^4 + y^4 + z^4}{4}\geq \sqrt[4]{1x^4y^4z^4} = |xyz|
$$
and equality is only obtained when $1 = x^4 = y^4 = z^4$. The only thing left is to take care of the sign of $xyz$, which will be positive iff all or exactly one of the variables is positive.
A: If $ x, y, z\ge 0$, then by AM-GM inequality we have $$ x^4+y^4+z^4+1\ge 4\sqrt[4]{x^4y^4z^4}=4xyz. $$
So the equality holds when $ x=y=z=1$. On the other hand, if one or three of the variables are negative, the RHS would be negative, but the LHS is always positive. Thus there are exactly two variables $< 0$. Wlog suppose that $ y, z <0$. Set $ u=-y $ and $ v=-z $. The equation becomes $ x^4+u^4+v^4+1=4xuv $. But this equation has the same form than the original one and we get that its solution is $ x=u=v=1$. Then $ y=z=-1$. 
Hence the solutions for the equation are $(x, y, z)=(1,1,1), (1-1,-1) $ and permutations.
