2 questions regarding solutions for $\sqrt{a+b} - (a-b)^2 = 0$

Question

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!

Answer 1

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.

Answer 2

We can rearrange to $$a+b=(a-b)^4$$

There are hence infinitely many solutions. For example, take $a+b=2^4=16$. Then $a-b=2$. We solve this to get $a=9, b=7$. Or take $a+b=3^4=81$. Then $a-b=3$. We solve this to get $a=42, b=39$.