4. Depending on the real parameter $a$, solve the equation

$$
x=a-\sqrt{a^{2}-x \sqrt{x^{2}+a^{2}}}
$$

in the set of real numbers.
4. The given equation is equivalent to $\sqrt{a^{2}-x \sqrt{x^{2}+a^{2}}}=a-x$. From this, we have the condition $a \geqslant x$, and after squaring the equation, it reduces to $a^{2}-x \sqrt{x^{2}+a^{2}}=a^{2}-2 a x+x^{2}$, i.e., $x \sqrt{x^{2}+a^{2}}=2 a x-x^{2}$. One possibility is $x=0$ (which is a solution whenever the condition $a \geqslant x$, i.e., $a \geqslant 0$, is satisfied); otherwise, after canceling with $x$, we are left with $\sqrt{x^{2}+a^{2}}=2 a-x$, and this, after squaring, with the condition $2 a-x \geqslant 0$, i.e., $a \geqslant \frac{x}{2}$, reduces to $x^{2}+a^{2}=4 a^{2}-4 a x+x^{2}$, i.e., $4 a x=3 a^{2}$. In the case $a=0$, this is always satisfied, i.e., the solutions are all numbers $x$ for which $a \geqslant x$ and $a \geqslant \frac{x}{2}$, i.e., $x \leqslant 0$. Now assume $a \neq 0$. Then from the last equation, we obtain another solution $x=\frac{3 a}{4}$; this is a solution of the original equation if the conditions $a \geqslant x$ and $a \geqslant \frac{x}{2}$, i.e., $a \geqslant \frac{3 a}{4}$ and $a \geqslant \frac{3 a}{8}$, are satisfied, which reduce to $\frac{a}{4} \geqslant 0$ and $\frac{5 a}{8} \geqslant 0$, i.e. (again) $a \geqslant 0$.

Thus, summarizing: for $a=0$ the solution of the equation is the entire interval $x \in(-\infty, 0]$, for $a>0$ the equation has two solutions, $x=0$ and $x=\frac{3 a}{4}$, and for $a<0$ the equation has no solutions.