4. Find all real numbers that satisfy the equation $\sqrt{x^{2}-p}+2 \sqrt{x^{2}-1}=x$, where $p$ is a parameter.
4. If we let $f(x)=\sqrt{x^{2}-p}+2 \sqrt{x^{2}-1}$, this problem then seeks the fixed point of $f(x)$, where $x$ must be greater than 0. If $px$, in this case, the original equation has no solution. Therefore, we can assume $p \geqslant 0$. From the original equation, we have $2 \sqrt{x^{2}-1}=x-\sqrt{x^{2}-p}$. Squaring and rearranging, we get $2 x^{2}+p-4=-2 x \sqrt{x^{2}-9}$. Squaring again and rearranging, we obtain $8(2-p) x^{2}=$ $(p-4)^{2}$. When $p=2$, this equation has no solution. When $p=4$, $x=0$ is not a solution to the original equation. Thus, we get $2-p>0$, meaning that in the case $0 \leqslant p<2$, there might be a solution $x=\frac{4-p}{\sqrt{8(2-p)}}$. Substituting this into the original equation for verification, we get $|3 p-4|=4-3 p$, which implies $3 p-4 \leqslant 0$. Therefore, only when $0 \leqslant p \leqslant \frac{4}{3}$ does the original equation have real roots, or fixed points, with the real root being $x=\frac{4-p}{\sqrt{8(2-p)}}$.