MHB How do I solve for non-negative integers x and y?

[anemone]

Here is this week's POTW:

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Solve for non-negative integers $x$ and $y$ of $\sqrt{xy}=\sqrt{x+y}+\sqrt{x}+\sqrt{y}$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to castor28 for his correct solution(Cool), which you can find below:

Spoiler

There is an obvious solution $x=y=0$; furthermore, as the radicals are positive, if $x=0$ then $y=0$ and conversely. We assume now that $x$ and $y$ are different from $0$.

We write the equation as:
$$
\sqrt{x+y} = \sqrt{xy} - \sqrt{x} - \sqrt{y}
$$
After squaring and simplifying, we get:
$$
xy = 2\sqrt{xy}(\sqrt{x} + \sqrt{y} - 1)
$$
and, since we assumed that $xy\ne0$,
\begin{align*}
\sqrt{xy}&=2(\sqrt{x}+\sqrt{y}-1)\\
(\sqrt{x}-2)(\sqrt{y}-2) &= 2
\end{align*}
This shows that $\sqrt{x}$ and $\sqrt{y}$ are both rational or both irrational. If they are irrational, they must belong to the same quadratic field; this implies that $y = q^2x$ for some rational number $q$.

Substituting in the equation, we get:
$$
qx - 2(q+1)\sqrt{x} + 2 = 0
$$

and this shows that the algebraic integers $\sqrt{x}$ and $\sqrt{y}$ are actually rational integers.

Writing $u=\sqrt{x}$ and $v=\sqrt{y}$, we get the equation:
$$
(u-2)(v-2) = 2
$$
to be solved in non-negative integers $u$ and $v$. One of the factors must be equal to $\pm1$ and the other must be equal to $\pm2$ (with the same sign). In terms of $u$ and $v$, this gives the solutions $(u,v)$ = $(3,4)$, $(4,3)$, $(0,1)$ and $(1,0)$, corresponding to $(x,y)$ = $(9,19)$, $(16,9)$, $(0,1)$ and $(1,0)$.

The first two solutions satisfy the original equation; the last two solutions are extraneous solutions introduced by the initial squaring.

To summarize, the only solutions in $(x,y)$ are $(0,0)$, $(9,16)$ and $(16,9)$.