3. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& x^{2}-y z=a, \\
& y^{2}-z x=a-1, \\
& z^{2}-x y=a+1
\end{aligned}
$$

with a real parameter $a$. Discuss the number of solutions.
Solution: If we subtract the second equation from the first and then the first equation from the third, we obtain the equalities

$$
(x-y)(x+y+z)=1 \quad \text { and } \quad(z-x)(x+y+z)=1 .
$$

It follows from them that the numbers $x-y$ and $z-x$ must be different from zero and both equal to the number $s=(x+y+z)$. Expressing $y=x-s$ and $z=x+s$ and substituting into the original system, we get:

$$
\begin{aligned}
& x^{2}-(x+s)(x-s)=a, \\
& (x-s)^{2}-x(x+s)=a-1, \\
& (x+s)^{2}-x(x-s)=a+1 .
\end{aligned}
$$

We easily find that this system is equivalent to the pair of equations $s^{2}=a$ and $3 x s=1$, from which, on the one hand, the solvability condition $a>0$ (since $s \neq 0$), and on the other hand, the expressions $s= \pm \sqrt{a}$ and $x=\frac{1}{3 s}= \pm \frac{1}{3 \sqrt{a}}$, so $y=x-s= \pm \frac{1-3 a}{3 \sqrt{a}}$ and $z=x+s= \pm \frac{1+3 a}{3 \sqrt{a}}$ (in all formulas, the same sign always applies). Since the system $(*)$ is solved by equivalent transformations, there is no need to perform a check.

Answer: For $a \leq 0$ the system has no solutions, for $a>0$ there are exactly two solutions $(x, y, z)$, namely

$$
\left(\frac{1}{3 \sqrt{a}}, \frac{1-3 a}{3 \sqrt{a}}, \frac{1+3 a}{3 \sqrt{a}}\right) \quad \text { and } \quad\left(\frac{-1}{3 \sqrt{a}}, \frac{-1+3 a}{3 \sqrt{a}}, \frac{-1-3 a}{3 \sqrt{a}}\right) .
$$