Show that $(\sqrt{y^2-x}-x)(\sqrt{x^2+y}-y)=y \iff x+y=0$

Let $x,y$ be real numbers such that $$\left(\sqrt{y^{2} - x\,\,}\, - x\right)\left(\sqrt{x^{2} + y\,\,}\, - y\right)=y$$ Show that $x+y=0$.

My try: Let $$\sqrt{y^2-x}-x=a,\sqrt{x^2+y}-y=b\Longrightarrow ab=y$$ and then $$\begin{cases} y^2=a^2+(2a+1)x+x^2\cdots\cdots (1)\\ x^2=b^2+(2b-1)y+y^2\cdots\cdots \end{cases}$$ $(1)+(2)$ then $$x=-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}\cdots\cdots (3)$$ so $$x+y=ab-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}=\dfrac{(a-b)(2ab-a+b)}{2a+1}$$ we take $(3)$ in $(2)$,we have $$b^2+(2b-1)y+y^2-x^2=\dfrac{(2ab-a+b)(2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b)}{(2a+1)^2}=0$$

so $$(2ab-a+b)=0$$ or $$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$ if $$2ab-a+b=0\Longrightarrow x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0$$ and if $$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$ I don't prove $$x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0?$$

Multiplication:

$y=x y-x \sqrt{x^2+y}-y \sqrt{-x+y^2}+\sqrt{x^2+y} \sqrt{-x+y^2} \Rightarrow$

$x \sqrt{x^2+y}+y \sqrt{-x+y^2}=\sqrt{x^2+y} \sqrt{-x+y^2}+xy-y$

squaring both sides:

$2 x y \sqrt{x^2+y} \sqrt{y^2-x}+x^4+x^2 y-x y^2+y^4= -x^3-x y+y^2-2 x y^2+2 x^2 y^2+y^3+(2 x y-2 y )\sqrt{x^2+y} \sqrt{-x+y^2} \Rightarrow $

simplifying:

$x^4+x^3-2 x^2 y^2+x^2 y+x y^2+x y+y^4-y^3-y^2 = -2 y \sqrt{x^2+y} \sqrt{y^2-x}$

Squaring again:

$y^8 - 2 y^7 - 4 x^2 y^6 + 2 x y^6 - y^6 + 6 x^2 y^5 + 2 y^5 + 6 x^4 y^4 - 2 x^3 y^4 + 3 x^2 y^4 - 4 x y^4 + y^4 - 6 x^4 y^3 - 4 x^3 y^3 - 2 x y^3 - 4 x^6 y^2 - 2 x^5 y^2 + x^4 y^2 + x^2 y^2 + 2 x^6 y + 4 x^5 y + 2 x^4 y + x^8 + 2 x^7 + x^6= -4 x^3 y^2+4 x^2 y^4-4 x y^3+4 y^5$

The only thing we have to check is how it factors as $(y+x) (p(x,y))$. Which gives us:

$(x+y)^2(x^6-2 x^5 y+2 x^5-x^4 y^2-2 x^4 y+x^4+4 x^3 y^3+2 x^3 y-x^2 y^4-4 x^2 y^3-4 x^2 y^2+2 x^2 y-2 x y^5+6 x y^4+2 x y^3+y^6-2 y^5-y^4-2 y^3+y^2)=0$

Assume that $x<0$ and $y>0$, your statement can be written (even if we change $x$ to $-x$),

If $x,y$ are strictly positive such that $(\sqrt{y^2+x}+x)(\sqrt{x^2+y}-y)=y$ then $x=y$.

This equality becomes, $$\overbrace{\dfrac{x+\sqrt{x+y^2}}{y+\sqrt{y+x^2}}}^{A}=\overbrace{\dfrac{y}{x^2+y-y^2}}^{B}.$$ Let us start with $x>y$, clearly $B<1$. Moreover $x\mapsto(x-\sqrt{y+x^2})$ is increasing and $x\mapsto \sqrt{y+x^2}$ strictly increasing, we conclude that $x\mapsto(x+\sqrt{x+y^2})-(y+\sqrt{y+x^2})$ is strictly increasing and thus $A>0$ when $x>y$. We deduces that $A>1$ and therefore $A$ cannot be equal to $B$. The case $x<y$ can be treated in the same way.

In the following proof we divide the $(x,y)$ plane into regions (see diagram) and show that each region can contain no solutions except on the line $x + y = 0$.

Let $$ F(x,y) = U(x,y)V(x,y) – y $$ where $$ U(x,y) = \sqrt{y^2 – x} \,\, – x \\V(x,y) = \sqrt{x^2 + y} \,\, – y $$

Then solutions satisfy $$ F(x,y) = 0 $$

Substituting $y=-x$ shows that $x+y=0$ is a solution for all x.

U is nonreal where $x>y^2$ (regions A and C in the diagram, bounded by red lines) and V is nonreal in the region $y<-x^2$ (regions B and C, also bounded by red lines). In these regions F is nonreal except possibly in region C, where U and V are both unreal. But there the condition for F to be real reduces to $x+y=0$, a subset of the known solution.

The following statements and deductions relate to the other regions of the $(x,y)$ plane, where $U$ and $V$ are real.

$U<0 \Leftrightarrow x>\tfrac{1}{2}(-1 + \sqrt{1 + 4y^2})$ (regions I, J).

$V<0 \Leftrightarrow y>\tfrac{1}{2}(1 + \sqrt{1 + 4x^2})$ (all regions except F, G).

$U_{x} < 0$ (regions D-K)

$V_{x} < 0 \Leftrightarrow x < 0$ (regions D, E, F)

$U_{xx} < 0$ (regions D-K)

$V_{xx} < 0 \Leftrightarrow y < 0$ (regions D, J, K)

where a subscript x denotes partial differentiation with respect to x.

On the diagram the lines on which $U=0$ and $V=0$ are coloured green and blue, respectively. It is easily shown that $F$ is nonzero on all the coloured lines (with sign as indicated) except at $(0,0)$ and $(1,-1)$. These lines delimit, but are excluded from, the regions A-K.

From the results above we can make the following deductions.

In region D: $$ U>0, V>0, U_{x}<0, V_{x}<0 \\F_{x} = UV_{x} + VU_{x} < 0 $$ This region is bounded on the right by the line $y<-x^2$, on which $F>0$. So $F>0$ throughout region D and it can contain no solutions.

In region E: $$ U>0, V>0, U_{x}<0, V_{x}<0 \\F_{x} < 0 $$ so here there can be no solutions other than those known to exist on the line $y=-x$.

In region F: $$ U>0, V<0, U_{x}<0, V_{x}<0, U_{xx}<0, V_{xx}>0 \\F_{xx} = UV_{xx} + VU_{xx} + 2U_{x}V_{x} > 0 $$ This region is bounded on the left by the line $V=0$ and on the right by the line $x = 0$, and on both these lines $F<0$. So the positive second derivative $F_{xx}$ means there can be no solutions $F=0$ in this region.

In a similar way, solutions can be ruled out for the following regions:

In region G, bounded on right by the line $V=0$ on which $F<0$: $$ U>0, V<0, U_{x}<0, V_{x}>0 \\F_{x}>0 $$

In region I, bounded on left by the line $U=0$ on which $F<0$: $$ U<0, V>0, U_{x}<0, V_{x}>0 \\F_{x} < 0 $$

In region J, containing a segment of the known solution line $x+y=0$ on which $F=0$: $$ U<0, V>0, U_{x}<0, V_{x}>0 \\F_{x} < 0 $$

In region K, bounded on the left and right by lines on which $F>0$: $$ U>0, V>0, U_{x}<0, V_{x}>0, U_{xx}<0, V_{xx}<0 \\F_{xx} = UV_{xx} + VU_{xx} + 2U_{x}V_{x} < 0 $$

Finally, in region H: $$ U>0, V>0, U_{x}<0, V_{x}>0 $$ and we note that $U_{x}<0$ in region G also, so for a given $y$, $U<U_{max}$, where $U_{max} = U(0,y) = y$

For the same value of y, $V<V_{max}$, where $V_{max} = V(X,y)$, and X is the value of x on the right-hand boundary of the region. On this boundary, $y=\sqrt{X^2+X}$, so $$ V_{max} = V(X,y) = \sqrt{X^2+y} \, – y < \sqrt{X^2+X+y} \, - y = \sqrt{y^2 + y} \,\, – y < \tfrac{1}{2}. $$

Therefore $$ F = UV – y < U_{max} V_{max} – y < y \tfrac{1}{2} – y = \, –\tfrac{1}{2} y < 0 $$ which completes the proof that there are no solutions other than $x+y = 0$.