3. If $a \geqslant 1$, then, when $\sqrt{a-\sqrt{a-x}}=x$, the solution is

保留源文本的换行和格式，直接输出翻译结果如下：

3. If $a \geqslant 1$, then, when $\sqrt{a-\sqrt{a-x}}=x$, the solution is
3. $\frac{\sqrt{4 a-3}-1}{2}$. Therefore, we get
$$
a^{2}-\left(2 x^{2}+1\right) a+\left(x^{4}-x\right)=0 .
$$

Solving, we get $a=\frac{\left(2 x^{2}+1\right) \pm(2 x+1)}{2}$,
which means $a=x^{2}+x+1$ or $a=x^{2}-x$.
From $a=x^{2}+x+1$ we get
$$
x=\frac{-1 \pm \sqrt{4 a-3}}{2}(\because a \geqslant 1, \therefore 4 a-3 \geqslant 0),
$$

which means $x=\frac{\sqrt{4 a-3}-1}{2}, x=-\frac{\sqrt{4 a-3}+1}{2}0$, a contradiction. Hence, $a=x^{2}-x$ has no solution.