TITLE: Find the number of triples $(x,y,z)$ of real numbers satisfying the equation $x^4+y^4+z^4+1=4xyz$
QUESTION [5 upvotes]: 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 :)

REPLY [3 votes]: 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.