ryan17 wrote: Determine all nonnegative integers $x, y$ satisfying the equation \begin{align*} \sqrt{xy}=\sqrt{x+y}+\sqrt{x}+\sqrt{y}. \end{align*} $x=0$ or $y=0$ imply solution $(x,y)=(0,0)$ $y\ne 0$ implies, setting $t=\frac xy$ : $x=(\sqrt{t+1}+\sqrt t+1)^2$ and $y=\frac xt$ with $t>0$ First equation implies $t=\left(\frac{x-2\sqrt x}{2\sqrt x-2}\right)^2$ with $x\ge 2$ and $x\ne 4$ And so $y=\frac{4(\sqrt x-1)^2}{(\sqrt x-2)^2}$ This easily implies $\sqrt x\in\mathbb N$ (the special case $x=y=2$ not being a solution) So $x=n^2$ and $y=\frac{4(n-1)^2}{(n-2)^2}$ and so $n-2\in\{1,2\}$ and the two solutions $(9,16)$ and $(16,9)$ Hence the answer $\boxed{(x,y)\in\{(0,0),(9,16),(16,9)\}}$

Previously posted here and here. (Solved by none other than pco at the first link!)

suppose: √x=z and √y=k then we have: zk=√(z^2 + k^2 ) + z+ k (zk-z-k)^2=z^2 + k^2 (z^2 k^2) +2zk=2zk(z+k) zk+2=2(z+k) (*) this shows that one of the z or k is even and another one is odd. suppose z is even then we can write from (*): z(k-2)+2=2k z=(2(k-1))/(k-2) we know that z is even so k-2 | k-1 and k is odd so k-1 is even and k-2 is odd so from this and the last line we have: 2k-4 | k-1 ---> 2k-4<=k-1 ---> k<=3 we can easily check these few k. hence the answer is:(0,0)-(9,16)-(16,9)

If we open it we get that xy=0 or xy=(x+y-(xy/4)-1)^2 and we understand in the second case we have gcd(x,y)=2^a where a<5 . Now we can easily find the answers