2020/012) Solve in real $x^4+y^4+z^4 + 1 = xyz$

We have LHS is positive so RHS is also positive.

So all of x,y,z are positive or one is positive and rest 2 are negative

So let us solve for positive x,y,z

Using AM GM inequality we have  $\frac{x^4+y^4+z^4+1}{4} >=\sqrt[4]{x^4y^4z^4}$

Or $x^4+y^4+z^4 +1 >= 4xyz$

And they are equal if $x=y=z$

So we get $3x^4-4x^3+1 = 0$

Or  we have trying $x=1$ and $x = 3$ x=1 is solution

giving one set of solution = $(1,1,1)$ and 2 negative  and one positive gives 3 more $(1,-1,-1), (-1,-1,1), \,and\, (-1,1,-1)$