Htykuut

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 $xmapsto(x-sqrt{y+x^2})$ is increasing and $xmapsto sqrt{y+x^2}$ strictly increasing, we conclude that $xmapsto(x+sqrt{x+y^2})-(y+sqrt{y+x^2})$ 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.

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:

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

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 $x+y = 0$.

Assuming continuity in the area interval $ (0 < x < 1 )$ and $ (0 > y > -1 ) $ would create problems as $x$ and $y$ are not always real in these areas.

Nay, union of inside parabola areas of $ y_1 = - x^2 $ and $ y_2 = sqrt{x} $ would violate $ x + y = 0, $ which is only the common chord of intersection of $ y_1,y_2$. So the shown line joining $(0,0)$ to $(1,-1)$ does not exist as real.
E.g., $(frac12, -frac12)$ does not lie on the
common line.

This is not a solution, but brute force can be used to remove the radicals. Let $A=y^2-x$ and $B=x^2+y$. We have

$$sqrt{AB}-ysqrt{A}-xsqrt{B}+xy=y$$

Isolating $sqrt{AB}$ and squaring both sides:

$$sqrt{AB}=ysqrt{A}+xsqrt{B}+y(1-x)quad(1)$$
$$AB=y^2A+x^2B+y^2(1-x)^2+2xysqrt{AB}+2y^2(1-x)sqrt{A}+2xy(1-x)sqrt{B}$$

(1) allows us to remove $sqrt{AB}$. We do this and also recall what $A$ and $B$ equal.

$$(y^2-x)(x^2+y)=y^2(y^2-x)+x^2(x^2+y)+y^2(1-x)^2+2xyleft(ysqrt{A}+xsqrt{B}+y(1-x)right)+2y^2(1-x)sqrt{A}+2xy(1-x)sqrt{B}$$

Group $sqrt{A}$ and $sqrt{B}$ terms, then rearrange a bit:

$$(y^2-x)(x^2+y)=y^2(y^2-x)+x^2(x^2+y)+y^2(1-x)^2+2xy^2(1-x)+2y^2sqrt{A}+2xysqrt{B}$$
$$x^2y^2+y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2xy^2+x^2y^2+2xy^2-2x^2y^2+2yleft(ysqrt{A}+xsqrt{B}right)$$
$$y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2x^2y^2+2yleft(ysqrt{A}+xsqrt{B}right)$$

(1) allows us to sub out the quantity in parentheses:

$$y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2x^2y^2+2yleft(y(x-1)+sqrt{AB}right)$$
$$y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2x^2y^2+2y^2(x-1)+2ysqrt{AB}$$
$$y^3-x^3-xy=y^4+xy^2+x^4+x^2y-y^2-2x^2y^2+2ysqrt{AB}$$
$$y^3-x^3-xy-y^4-xy^2-x^4-x^2y+y^2+2x^2y^2=2ysqrt{AB}$$

Squaring both sides, we've reached a goal of no longer having radicals.

$$(y^3-x^3-xy-y^4-xy^2-x^4-x^2y+y^2+2x^2y^2)^2=4y^2(y^2-x)(x^2+y)$$

I had a CAS expand this, move it all to one side, and then, as expected, $(x+y)$ factors out of it (twice).

$$(x+y)^2 p(x,y)=0$$

where $$p(x,y)=x^6-2 x^5 y+2 x^5-x^4 y^2-2 x^4 y+x^4+4 x^3 y^3+2 x^3 y-x^2 y^4-4 x^2 y^3-4 x^2 y^2+2 x^2 y-2 x y^5+6 x y^4+2 x y^3+y^6-2 y^5-y^4-2 y^3+y^2$$

is a monster. It would be sufficient to show that $p(x,y)$ is never $0$ in the region of the plane where both $sqrt{A}$ and $sqrt{B}$ are defined aside from points along $x+y=0$ (like $(0,0)$). This is a pretty messy polynomial, but at least it's a polynomial.

EDIT: This approach seems to be useless; a CAS plot of the zero set of $p$ has several components, all of which are in the region where $sqrt{A}$ and $sqrt{B}$ are defined. They must be extraneous solutions from the squaring that was done twice.

A more generalised approach over my earlier post. This is not intended to be an exhaustive proof but an experimental one. Constructive comments are most welcome.

Let

$$sqrt{y^2-x}-x=Ay^nqquad cdots (1)\
sqrt{x^2+y}-y=frac {y^{1-n}}A qquad cdots (2)\$$
such that the original equation $$left(sqrt{y^2-x}-xright)left(sqrt{x^2+y}-yright)=y$$
is satisfied as required.

From $(1)$,

$$begin{align}
sqrt{y^2-x}&=x+Ay^n\
y^2-x&=x^2++2Axy^n+A^2y^{2n}\
y^{2n}A^2+2xy^nA+(x^2-y^2+x)&=0\
A^2+frac {2x}{y^n}A+frac{(x^2-y^2+x)}{y^{2n}}&=0qquad qquad qquad qquad cdots (3)
end{align}$$

From $(2)$,

$$begin{align}
sqrt{x^2+y}&=y+frac {y^{1-n}}A\
Asqrt{x^2+y}&=Ay+y^{1-n}\
A^2(x^2+y)&=A^2y^2+2Ay^{2-n}+y^{2(1-n)}\
(x^2-y^2+y)A^2-2y^{2-n}A-y^{2(1-n)}&=0\
A^2-frac{2y^{2-n}}{x^2-y^2+y}A-frac{y^{2(1-n)}}{x^2-y^2+y}&=0qquad qquad qquad cdots (4)
end{align}$$

Equating coefficients of $A^1$:
$$begin{align}frac{2x}{y^n}&=-frac{2y^{2-n}}{x^2-y^2+y}\
y^2&=-x(x^2-y^2+y)qquad qquad qquad qquad qquad qquad cdots (5)end{align}$$

Equating coefficients of $A^0$:

$$begin{align}frac{x^2-y^2+x}{y^{2n}}&=-frac{y^{2(1-n)}}{x^2-y^2+y}\
y^2&=-(x^2-y^2+x)(x^2-y^2+y)qquad cdots (6)end{align}$$

(5)=(6):
$$begin{align}x(x^2-y^2+y)&=(x^2-y^2+x)(x^2-y^2+y)\
(x^2-y^2)(x^2+y^2-y)&=0\
(x-y)(x+y)(x^2-y^2+y)&=0\
Rightarrow x-y=&0, x+y=0, x^2-y^2+y=0end{align}$$

Checking by substitution into the original equation shows that only
$$x+y=0$$
is valid.

This graph created on desmos.com might help illustrate the approach:

https://www.desmos.com/calculator/qrlbgbalix

We need to prove that $x=y$, where
$$left(sqrt{y^{2}+x}+xright)left(sqrt{x^{2} + y}- yright)=y$$ or

$$left(sqrt{y^{2}+x}+xright)left(sqrt{x^{2} + y}- yright)=left(sqrt{x^{2}+y}+xright)left(sqrt{x^{2} + y}- xright)$$ or
$$left(sqrt{y^{2}+x}+xright)left(sqrt{x^{2} + y}- yright)-left(sqrt{x^{2}+y}+xright)left(sqrt{x^{2} + y}- yright)+$$
$$+left(sqrt{x^{2}+y}+xright)left(sqrt{x^{2} + y}- yright)-left(sqrt{x^{2}+y}+xright)left(sqrt{x^{2} + y}- xright)=0$$ or
$$left(sqrt{y^{2}+x}-sqrt{x^2+y}right)left(sqrt{x^{2} + y}- yright)+left(sqrt{x^{2}+y}+xright)(x-y)=0,$$ which gives $x=y$ or

$$frac{(1-x-y)left(sqrt{x^{2} + y}- yright)}{sqrt{y^{2}+x}+sqrt{x^2+y}}+sqrt{x^{2}+y}+x=0,$$ which is
$$sqrt{x^2+y}-ysqrt{x^2+y}+sqrt{(x^2+y)(y^2+x)}+xsqrt{y^2+x}+x^2+xy+y^2=0.$$
Now, we'll consider four cases.

1. 
   $xgeq0$, $ygeq 0$.

Since $$-ysqrt{x^2+y}+sqrt{(x^2+y)(y^2+x)}=sqrt{x^2+y}left(sqrt{y^2+x}-yright)geq0,$$ we obtain $x=y=0.$

2. 
   $xgeq0,$ $yleq0.$

It's obvious that this case gives $x=y=0$ again.

3. 
   $xleq0$, $ygeq0.$
   

Since, $$sqrt{(x^2+y)(y^2+x)}+xsqrt{y^2+x}=sqrt{y^2+x}left(sqrt{x^2+y}+xright)geq0,$$ it's enough to prove that
$$x^2+xy+y^2geq(y-1)sqrt{x^2+y},$$ which is obvious for $yleq1.$

But for $ygeq1$ by AM-GM we obtain:
$$(y-1)sqrt{x^2+y}leqfrac{1}{2}((y-1)^2+x^2+y)$$ and it's enough to prove that
$$x^2+xy+y^2geqfrac{1}{2}((y-1)^2+x^2+y)$$ or
$$require{cancel} cancel{(x+y)^2+y^2+y-1geq0.}\
(x+y)^2+y-1geq0.$$
We see that for $ygeq1$ the equality does not occur and in the case $y<1$ the equality occurs for
$$x^2+xy+y^2=(y-1)sqrt{x^2+y}=0,$$ which gives $x=y=0$ again.

4. 
   $xleq0$ and $yleq0.$
   

In this case it's enough to prove that
$$xy+xsqrt{y^2+x}geq0$$ or
$$xleft(y+sqrt{y^2+x}right)geq0,$$ which is obvious.

The equality occurs for $x^2+y^2=0$ and we got $x=y=0$ again.

Done!