Consider the following
For all $a\in\Bbb R$ solve the equation $$\sqrt{x^2+4a^2\sqrt{x+a}}=x+2a$$
It is immediate to see that we got the restriction $x\geqslant-a$ (even though not given I assume that this equation is meant only for $x\in\Bbb R$ aswell). However, squaring twice gives \begin{align*} \sqrt{x^2+4a^2\sqrt{x+a}}&=x+2a\\ x^2+4a^2\sqrt{x+a}&=x^2+4a^2+4ax\\ a\sqrt{x+a}&=x+a\\ a^2(x+a)&=(x+a)^2\tag1\\ a^2&=x+a\tag2 \end{align*} From $(1)$ we get that $x_1=-a$ is a possible solution and assuming that $x\neq-a$ we may divide by $(x+a)$ to obtain $x_2=a(a-1)$ as second possibility.
Hence we applied a non-equivalent transformation we are forced to check our derived solutions. Thus, we get for $x_1=-a$ \begin{align*} \sqrt{(-a)^2+4a^2\sqrt{-a+a}}&=-a+2a\\ \sqrt{a^2}&=a\\ |a|&=a \end{align*} Which only holds for $a\geqslant0$. For $x_2=a(a-1)$ we get \begin{align*} \sqrt{(a(a-1))^2+4a^2\sqrt{a(a-1)+a}}&=a(a-1)+2a\\ \sqrt{a^4-2a^3+a^2+4a^2|a|}&=a(a+1) \end{align*} First we may assume $a\geqslant0$ again from where we conclude \begin{align*} \sqrt{a^4-2a^3+a^2+4a^3}&=a(a+1)\\ \sqrt{a^4+2a^3+a^2}&=a(a+1)\\ \sqrt{a^2(a+1)^2}&=a(a+1)\\ |a(a+1)|&=a(a+1) \end{align*} The last line holds since $a\geqslant0\implies a(a+1)\geqslant0$. On the other hand from $a<0$ we get \begin{align*} \sqrt{a^4-2a^3+a^2-4a^3}&=a(a+1)\\ \sqrt{a^4-6a^3+a^2}&=a(a+1)\\ |a|\sqrt{a^2-6a+1}&=a(a+1)\\ \sqrt{a^2-6a+1}&=-a-1 \end{align*} But I am not sure how to interpret this result. Apparently it does not hold for $a=-1$, which matches the given restriction, but I am not sure whether this is sufficient to show that this is a contradiction.
I was told that the solution is given by $x=-a$ and $a=0$. This would basically mean that there is only the trivial solution $(x,a)=(0,0)$. However, since I cannot spot a mistake above I am certainly sure that the second restriction, $a=0$, is nonsense and I do not see how it could be derived. Moreover if I am not mistaken the pair $(x,a)=(-5,5)$ works perfectly well too (choosing the positive square root of $25$). I thought about somehow combining $x\geqslant-a$ and $a\geqslant0$ to obtain $a=0$ but I do not see how.
Could someone explain me what I am missing? Is my solution correct or is it in need of improvement?
Thanks in advance!
Squaring the equation (which might introduce extra solutions), we have
$$x^2+4a^2\sqrt{x+a}=x^2+4ax+4a^2,$$ or $$a=0\lor a\sqrt{x+a}=x+a.$$
$a=0$ reduces the equation to $|x|=x$, or $x\ge0$.
Otherwise, as $x+a\ge0$, a solution is only possible with $a>0$ and is given by
$$\sqrt{x+a}=0\lor\sqrt{x+a}=a,$$ i.e.
$$x=-a\lor x=a^2-a.$$ If you plug these values in the original equation, they are compatible.
To summarize:
$a<0$: no solution,
$a=0$: $x\ge0$,
$a>0$: $x=-a$ or $x=a^2-a$.