Why, after squaring, is the smallest root of the quartic the solution of the real problem? Looking at this post, I considered the more general problem of the solution of equation $$x+\sqrt{x+\sqrt{a+x}}=a\tag 1$$ where $a> 0$ .
Using successive squaring processes, this leads to the quartic 
$$x^4-2 (2
   a+1) x^3+\left(6 a^2+4 a+1\right) x^2-\left(4 a^3+2 a^2+1\right) x+a \left(a^3-1\right)=0\tag2$$ the discriminant of which being
$$\Delta=1024 a^4-1280 a^3-768 a^2-80 a-23$$ which is positive for any $a>\sqrt 3$; then, using the notations used here, since $P=-4(4a+1)$ and $D=-32(4a+1)$, all four roots are real and distinct. Extensive numerical tests showed that all roots are positive.

How could be explained the fact that the only solution of $(1)$ systematically corresponds to the smallest root of $(2)$ ? 
How could be explained the fact that, for a few integer values of $a$, one of the solution of $(2)$ is also an integer ? Testing for the range $2 \leq a \leq 1000$, this is the case for the values $a=2,3,5,11,46,68,250,296,566,636$.

 A: Squarising, let A is equation, B is condition
$A:x+\sqrt{a+x}=(a-x)^2,B_1:a>x$
$⇔A:a+x=((a-x)^2-x)^2,B_2:(a-x)^2-x\geq0,B_3: x\geq -a$
$⇔A:$ your equation, $|x|\leq a, x=\dfrac{(2a+1)±\sqrt{4a+1}}2,B_4:a\geq-\dfrac{1}4, $
This equation needs $|x|\leq a$ from $B_1,B_3$  and  $ 0<\dfrac{2a+1-\sqrt{4a+1}}2\leq x\leq a<\dfrac{2a+1+\sqrt{4a+1}}2 $.
For example $a=11$, $8≒\dfrac{23-\sqrt{45}}2\leq x\leq11<\dfrac{23+\sqrt{45}}2$ 
Thus this equation holds between relatively small x. The extra equations which we get by squarising have negative sign of x.
But since LHD of first equation are all positive value, LHD reach to RHD in shortest distance by other equations. 
A: If the real and positive solutions from the 4th-order equations are $x_1 < x_2 < x_3 < x_4$ and we introduce
$$
y_i^{(\alpha,\beta)} = x_i + \alpha \sqrt{x_i + \beta \sqrt{x_i + a}}
$$
where $\alpha,\beta = \pm$ and only take the positive roots. For each $(\alpha,\beta)$ there is a unique $x_i$ that will result in $y_i^{(\alpha,\beta)}=a$. 
We can now make two observations:


*

*From the ordered $x_i$ it automatically follows that:
$$
y_1^{(++)} < y_2^{(++)} <y_3^{(++)} <y_4^{(++)}
$$ 

*For the different possibilities of $x_1$ we have
$$
\max \left( y_1^{(--)}, y_1^{(-+)}, y_1^{(+-)}, y_1^{(++)} \right) = y_1^{(++)} 
$$
Hence if the solution of $x + \sqrt{x + \sqrt{x+ a}} =a$ would not correspond to $x_1$ (the smallest) but to $x_j$ one would obtain that $y_1^{(\alpha,\beta)} \leq y_1^{(++)} < a=y_j$, which contradicts the fact that there should be such a combination $(\alpha,\beta)$.
With respect to the solutions you missed the case $a=250$ and $a=296$ for values $a \leq 500$.
We have the following set of equations in integers at the different levels of square roots:
$$
x + a = n^2
$$
$$
x + n = m^2
$$
$$
x + m = a
$$
for some integers $n,m \in \mathbb{Z}$. From the second we obtain $n=m^2-x$ and substitution of this and the third equation in the first gives:
$$
x + (x+m) = (m^2 - x)^2
$$
and hence
$$
x^2 - 2(m^2+1)x + m(m^3 -1) = 0
$$
With two solutions:
$$
x_\pm = (m^2+1) \pm \sqrt{2 m^2 + m + 1}
$$
and the corresponding values for $a$ are:
$$
a_\pm = (m^2+m+1) \pm \sqrt{2 m^2 + m + 1}
$$
In order for this to be integers we only need to find value of $m$ (positive and negative) such that $2 m^2 + m + 1$ is a perfect square.
So the problem simplifies to finding integer solutions of the equation:
$$
2 m^2 + m + 1 = p^2
$$
This can be rewritten as
$$
8 p^2 - (4 m + 1)^2 = 7
$$
which is a type of Pell equation. The succesive solutions of $m_i$ that results in such a perfect square are given by $m_0=0$, $m_1=1$ and the recurrence relation $m_{i+1}=-6 m_{i} - m_{i-1} - 2$. Note that in order to obtain all solutions $(x,a)$ of the problem above, both positive and negative values of $m$ are allowed and that the recurrence relation needs to be followed in both directions. 
With the substitution $m_i \rightarrow \frac{1}{4} (u_i - 1)$, a normal recurrence relation is obtained with $u_0=0$, $u_1=5$, and $u_{i+1}=-6 u_{i} - u_{i-1}$. Working this out properly gives:
$$
m_i = \frac{1 - 2 \sqrt{2}}{8} \left( -3 - 2 \sqrt{2} \right)^i + \frac{1 + 2 \sqrt{2}}{8} \left( -3 + 2 \sqrt{2} \right)^i - \frac{1}{4}
$$
and results in 
$$
\begin{array}{llllll}
m_0=0 & m_1=1     & m_2=-8    & m_3=45     & m_4=-264   & \dots \\
      & m_{-1}=-3 & m_{-2}=16 & m_{-3}=-95 & m_{-4}=552 & \dots
\end{array}
$$
Inserting this sequence in the expressions for the solutions $(x,a)$ gives:
$$
(x_+,a_+)=\dots,(280,296),(14,11),(2,2),(4,5),(76,68),\dots
$$
$$
(x_-,a_-)=\dots,(234,250),(6,3),(0,0),(0,1),(54,46),\dots
$$
which are all integer combinations for which $x + a  - ((a-x)^2 - x)^2 == 0$. Only a subset of them will also give $x + \sqrt{x + \sqrt{x+ a}} =a$. 
Note that also the $(x_i,a_i)$ could be written in a similar form, but more terms, as done for $m_i$.
A: Ad (1): An equation $\Psi(x)=0$ (together with explaining text) defines a solution set $S$ in an implicit way. For any given $x$ it is easy to decide whether $x\in S$ or not; but what we really want is an explicit description of $S$ in the form of a list, a parametric representation, or similar. In order to arrive at such a list we perform  a chain of arguments like
$$\Psi(x)=0\ \Rightarrow \ \ldots\ \Rightarrow \ \ldots\ \Rightarrow x\in\tilde S\ ,$$
where $\tilde S$ is an explicitly presented set; in your example a set of four real numbers. In this way we have not proven $S=\tilde S$, but only $S\subset\tilde S$, and we now have to check which elements of $\tilde S$ are in fact solutions of the original problem. In your example you have introduced spurious solutions by repeated squaring of the given equation.
Ad (2): This is numerology.
