Q34 / 90 · KVPY 2016 (SB/SX) · Mathematics

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

  • A
    0
  • Correct answer
    B
    4
  • C
    8
  • D
    more than 8

Solution

$(x, y, z)$ are real & $x^{4}, y^{4}, z^{4}$ are positive real numbers $$ \begin{array}{l} \therefore \frac{x^{4}+y^{4}+z^{4}+1}{4} \geq|x y z| \\ \Rightarrow(x y z) \geq|x y z| \\ \text { i.e., } x y z>0 \end{array} $$ So it holds equality $\therefore \mathrm{x}^{4}=\mathrm{y}^{4}=\mathrm{z}^{4}=1 ;$ But $\mathrm{xyz}>0$ $\therefore(\mathrm{x}, \mathrm{y}, \mathrm{z}) \in\{(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)\}$ So no. of triplets is 4 .

Hence, the correct answer is option: 'B'.

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