4. If $\left\{\begin{array}{l}x^{2}-y z=1 \\ y^{2}-x z=1 \\ z^{2}-x y=1\end{array}\right.$ has solutions $(x, y, z)$ (where $x, y, z$ are rational numbers) the number of solution sets is ( )
A. No solution
B. 3
C. 6
D. Infinite sets
4. C

So $x=y$ or $x+y=z$. For (2)-(3) there is a similar conclusion. Also, if $x=y=z$, then $x^{2}-yz=0$ which is a contradiction. Without loss of generality, assume $x \neq y$, then one of $x, y$ must be equal to $z$, otherwise, $\left\{\begin{array}{l}x+y=z \\ y+z=x, \\ x+z=y\end{array}\right.$ which leads to $x=y=z=0$. This results in $x^{2}-yz=0$, a contradiction. When $x=z \neq y$, the solutions are $(1,0,1),(-1,0,-1)$. When $y=z \neq x$, the solutions are $(0,1,1)(0,-1,-1)$. When $x=y$, we have $x=y \neq z$, the solutions are $(1,1,0)(-1,-1,0)$. Therefore, there are 6 solutions in total, so the answer is C.