A6 (5-1, Czechoslovakia) Find all real solutions of the equation $\sqrt{x^{2}-p}+2 \sqrt{x^{2}-1}=x$, where $p$ is a real parameter.
Given that $px$, the original equation has no solution, so we can assume $p \geqslant 0$, and $x \geqslant 1, x \geqslant p$.
Now, we rewrite the equation in the form
$$
2 \sqrt{x^{2}-1}=x-\sqrt{x^{2}-p},
$$

Squaring and rearranging, we get
$$
2 x^{2}+p-4=-2 x \sqrt{x^{2}-p} .
$$

Squaring again and rearranging, we get
$$
8(2-p) x^{2}=(p-4)^{2} .
$$

It is clear that $p$ must satisfy $0 \leqslant p<2$. And in this case, the only possible solution is
$$
x=\frac{4-p}{\sqrt{8(2-p)}} .
$$

Substituting this into the original equation and simplifying, we get
$$
|3 p-4|=4-3 p,
$$

Thus, we have $p \leqslant \frac{4}{3}$.
In summary, the equation has a unique solution $x=\frac{4-p}{\sqrt{8(2-p)}}$ if and only if $0 \leqslant p \leqslant \frac{4}{3}$.