2 questions regarding solutions for $\sqrt{a+b} - (a-b)^2 = 0$ Here's two questions derived from the following question:
$\quad\begin{matrix}
\text{Is there more than one solution to the following statement?} \\
\!\sqrt{a+b} - (a-b)^2 = 0 
\end{matrix}$
$\color{Blue}{(1)\!\!:\;}$How would one (dis)prove this? I.e. In what ways could one effectively determine whether an equation has more than one solution; More specifically, this one?
$\color{Blue}{(2)\!\!:\;}$Is it possible to determine this with(out) a valid solution as a sort of reference? 
Cheers!

If you are looking for real solutions, then note that $a+b$ and $a-b$ are just arbitrary numbers, with $a + b \ge 0$. This is because the system
$$
\begin{cases}
u = a + b\\
v = a - b
\end{cases}
$$
has a unique solution
\begin{cases}
a = \frac{u+v}{2}\\
\\
b = \frac{u-v}{2}.
\end{cases}
In the variables $u,v$ the general solution is $u = v^{4}$, which translates to
$$
a = \frac{v^{4}+v}{2}, \qquad
b = \frac{v^{4}-v}{2},
$$
as noted by Peter Košinár. Note that $a+b = u = v^{4}$ is always non-negative, as requested.
If it's integer solutions you're looking for, you get the same solutions (for $v$ an integer), as $v$ and $v^{4}$ have the same parity.
