Question

The number of triples $(x,y,z)$ of real numbers satisfying the equation
$x^4+y^4+z^4+1=4xyz$ is

• A. $0$
• B. $4$
• C. $8$
• D. More than $8$

Answer 1

The correct option is B $4$

Applying $AM\ge GM$

We get $\frac{x^4+y^4+z^4+1}{4}\ge \sqrt[4]{4x^4{y}^4{z}^4}$

$⟹x^4+y^4+z^4+1\ge 4xyz$

$\text{Thus, min value of}{x}^4+y^4+z^4+1\text{must be}4xyz$

Clearly our function is increasing for x,y,z>0 and decreasing for x,y,z<0.

Thus min value must occur for integral values like 1,0,-1.

For 0, it doesn't satisfy the equation $x^4+y^4+z^4+1=4xyz$

But triplets of 1 and -1 do.

Thus, we can check that the triplets $(1,-1,-1);(-1,1,-1);(-1,-1,1)and(1,1,1)$ satisfy the equation.

We don't need to look for further values as we won't get min value of $x^4+y^4+z^4+1$ by them.

Thus, these are the only solutions.