The problem is:
For real numbers x and y, if:
$(\sqrt{y^2+x}+x)(\sqrt{x^2+y}-y)=y$
then $x=y$.
Firstly, I prove that for $y=0$, we have $x=0$.
Then I make $x=ky$ for real $k$.
=>$(\sqrt{y^2+ky}+ky)(\sqrt{k^2y^2+y}-y)=y$
<=> $(\sqrt{y+k}+k\sqrt{y})(\sqrt{k^2y+1}-\sqrt{y})=1$
I want to get k=1 but found no way to get there.
Any help would be appreciated!
We introduce the notations: $$ \begin{aligned} a &=\sqrt{y^2 +x}\ge 0\ ,\text{ so}&a^2 &=y^2 + x\ ,\\ b &=\sqrt{x^2 +y}\ge 0\ ,\text{ so}&b^2 &=x^2 + y\ ,\\ &\qquad\text{ and there is the relation}\\ y&=(a+x)(b-y)\ . \end{aligned} $$ The idea is to eliminate $x,y$, and obtain a single equation connecting the variables $a,b$.
We extract $x=a^2 -y^2$ from the first equation, insert in the other two, and obtain: $$ \left\{ \begin{aligned} 0 &=(y^2 - a^2)^2-b^2+y\ ,\\ 0 &=(y^2-a^2-a)(y-b)-y\ , \end{aligned} \right. \qquad\text{i.e}\qquad \left\{ \begin{aligned} 0 &= y^4 -2a^2 y^2 + a^4-b^2+y\ ,\\ 0 &= y^3 -by^2 -(a^2+a+1)y +(a^2+a)b\ . \end{aligned} \right. $$ We need to eliminate $y$ between the two polynomials. We can successively apply the euclidean algorithm, so divide first the $Y$-polynomial $(Y^2 - a^2)^2+Y-b^2$ by the quotient $(Y^2-a^2-a)(Y-b)-Y$, and obtain a rest of degree $<3$, then repeat the process with the last two polynomials (quotient and rest). This is the same as computing a resultant of two polynomials. In our case, the determinant $\Delta$ (which is the resultant) seems to be slightly to big for a display, but many entries vanish.
We know that this resultant $\Delta$ is zero because the polynomials have a common root $y$!
Before we start the computation, there is one observation that we must make. The resultant is a polynomial in $a,b$. If in this resultant we set $b=a$, we become zero. (Why? Because the two polynomials we start with, when setting $b=a$ in them, become $(y^2-a^2)^2-a^2+y=((y^2-a^2)+y)((y^2-a^2)-y+1)$, and $(y^2-a^2-a)(y-a)-y=(y^2-a^2)(y-a)-ay+a^2-y = ((y^2-a^2)+y)((y-a)-1)$. And the two polynomials share now a common factor, $y^2-a^2+y$, as expected. So we should be in position to factor sooner or later $(a-b)$. This is deceptively simple for a computer, but i am trying to also give the human way.
Pivots are boxed. $$ \small \begin{aligned} \Delta &= \begin{vmatrix} \boxed 1 & 0 & -2a^2 & 1 & a^4-b^2\\ & \boxed 1 & 0 & -2a^2 & 1 & a^4-b^2\\ & & 1 & 0 & -2a^2 & 1 & a^4-b^2\\ 1 & -b & -(a^2 + a + 1) & (a+1)ab\\ & 1 & -b & -(a^2 + a + 1) & (a+1)ab\\ & & 1 & -b & -(a^2 + a + 1) & (a+1)ab\\ & & & 1 & -b & -(a^2 + a + 1) & (a+1)ab \end{vmatrix} \\[3mm] &= \begin{vmatrix} 1 & 0 & -2 a^{2} & 1 & a^{4} - b^{2} & 0 & 0 \\ 0 & 1 & 0 & -2 a^{2} & 1 & a^{4} - b^{2} & 0 \\ 0 & 0 & 1 & 0 & -2 a^{2} & 1 & a^{4} - b^{2} \\ 0 & 0 & a^{2} - a - 1 & -a^{2} b + a b - 1 & -a^{4} + b^{2} + b & a^{4} b - b^{3} & 0 \\ 0 & 0 & -b & a^{2} - a - 1 & a^{2} b + a b - 1 & -a^{4} + b^{2} & 0 \\ 0 & 0 & 1 & -b & -a^{2} - a - 1 & a^{2} b + a b & 0 \\ 0 & 0 & 0 & 1 & -b & -a^{2} - a - 1 & a^{2} b + a b \end{vmatrix} \\[3mm] &= \begin{vmatrix} {}\boxed 1 & 0 & -2 a^{2} & 1 & a^{4} - b^{2} \\ a^{2} - a - 1 & -a^{2} b + a b - 1 & -a^{4} + b^{2} + b & a^{4} b - b^{3} & 0 \\ -b & a^{2} - a - 1 & a^{2} b + a b - 1 & -a^{4} + b^{2} & 0 \\ 1 & -b & -a^{2} - a - 1 & a^{2} b + a b & 0 \\ 0 & 1 & -b & -a^{2} - a - 1 & a^{2} b + a b \end{vmatrix} \\[3mm] &= \begin{vmatrix} {}\boxed 1 & 0 & -2 a^{2} & 1 & a^{4} - b^{2} \\ a^{2} - a - 1 & -a^{2} b + a b - 1 & -a^{4} + b^{2} + b & a^{4} b - b^{3} & 0 \\ 0 & a^{2} - a - 1 & -a^{2} b + a b - 1 & -a^{4} + b^{2} + b & a^{4} b - b^{3} \\ 0 & -b & a^{2} - a - 1 & a^{2} b + a b - 1 & -a^{4} + b^{2} \\ 0 & 1 & -b & -a^{2} - a - 1 & a^{2} b + a b \end{vmatrix} \\[3mm] &= \begin{vmatrix} -a^{2} b + a b - 1 & a^{4} - 2 a^{3} - 2 a^{2} + b^{2} + b & a^{4} b - b^{3} - a^{2} + a + 1 & -(a^2 - b) \cdot (a^{2} + b) \cdot (a^{2} - a - 1) \\ a^{2} - a - 1 & -a^{2} b + a b - 1 & -a^{4} + b^{2} + b & b(a^2-b)(a^2+b) \\ -b & a^{2} - a - 1 & a^{2} b + a b - 1 & \boxed{-(a^2-b)(a^2+b)} \\ 1 & -b & -a^{2} - a - 1 & ab(a+1) \end{vmatrix} \\[3mm] &= \begin{vmatrix} -b - 1 & -a^{2} + b^{2} - 2 a + b - 1 & 2 a^{2} b - b^{3} + a b & 0 \\ a^{2} - b^{2} - a - 1 & -b - 1 & -a^{4} + a^{2} b^{2} + a b^{2} + b^{2} & 0 \\ -b & a^{2} - a - 1 & a^{2} b + a b - 1 & -(a^2-b)(a^2+b) \\ \boxed 1 & -b & -a^{2} - a - 1 & ab(a+1) \end{vmatrix} \\[3mm] &\qquad\text{ (From the second row subtract the first one)} \\[3mm] &= \begin{vmatrix} -b - 1 & -a^{2} + b^{2} - 2 a + b - 1 & 2 a^{2} b - b^{3} + a b & 0 \\ (a-b)(a+b+1) & (a-b)(a+b+2) & -(a-b)(a^3 + a^2b + 2ab + b^2 + b) & 0 \\ -b & a^{2} - a - 1 & a^{2} b + a b - 1 & -(a^2-b)(a^2+b) \\ 1 & -b & -a^{2} - a - 1 & ab(a+1) \end{vmatrix} \\[3mm] &= (a-b) \begin{vmatrix} -b - 1 & -a^{2} + b^{2} - 2 a + b - 1 & 2 a^{2} b - b^{3} + a b & 0 \\ a + b - 1 & a + b + 2 & -a^{3} - a^{2} b - 2 a b - b^{2} - b & 0 \\ -b & a^{2} - a - 1 & a^{2} b + a b - 1 & -(a^2-b)(a^2+b) \\ \boxed 1 & -b & -a^{2} - a - 1 & ab(a+1) \end{vmatrix} \\[3mm] &= (a-b) \begin{vmatrix} -b - 1 & -(a+1)^2 & a^{2} b - b^{3} - a^{2} - a - b - 1 & 0 \\ a + b - 1 & a b + b^{2} + a + 2 & -a b - b^{2} - 1 & 0 \\ -b & a^{2} - b^{2} - a - 1 & -b - 1 & -a^4 + b^2 \\ 1 & 0 & 0 & ab(a+1) \end{vmatrix} \\[3mm] &\qquad\text{ Multiply last row by $(a-1)$, add it to previous one} \\[3mm] &= (a-b) \begin{vmatrix} -b - 1 & -(a+1)^2 & a^{2} b - b^{3} - a^{2} - a - b - 1 & 0 \\ a + b - 1 & a b + b^{2} + a + 2 & -a b - b^{2} - 1 & 0 \\ a - b - 1 & a^{2} - b^{2} - a - 1 & -b - 1 & -(a-b)(a^3+b) \\ \color{red}{1} & 0 & 0 & \color{red}{ab(a+1)} \end{vmatrix} \\[3mm] &=\color{red}{1}\cdot(a-b)^2(a^3+b) \begin{vmatrix} -(a+1)^2 & a^{2} b - b^{3} - a^{2} - a - b - 1 \\ a b + b^{2} + a + 2 & -a b - b^{2} - 1 \end{vmatrix} \\ &\qquad +\color{red}{ab(a+1)}(a-b) \begin{vmatrix} -b - 1 & -(a+1)^2 & a^{2} b - b^{3} - a^{2} - a - b - 1 \\ a + b - 1 & a b + b^{2} + a + 2 & -a b - b^{2} - 1 \\ a - b - 1 & a^{2} - b^{2} - a - 1 & -b - 1 \end{vmatrix} \\[3mm] &=\dots \end{aligned} $$ It turns out that from the $3\times3$-determinant we can further extract one more factor $(a-b)$. The final result is: $$ \begin{aligned} \Delta =(a-b)^2&\Big( \ (a-b)^2(a+b)^4 \\ &\qquad\qquad +(a + b) (4 a^{4} + 4 a^{3} b + 5 a^{3} + a^{2} b + 3 a b^{2} + 3 b^{3}) \\ &\qquad\qquad\qquad\qquad +(a + b) (3 a^2 + a b + 2 b^2) \\ &\qquad\qquad\qquad\qquad\qquad\qquad +5ab + 2b^2 + 3b\Big)\ . \end{aligned} $$ Recall that $a,b\ge 0$.
This quantity $\Delta$ is $\ge 0$, and can become zero only if $a=b$.
From here we obtain $$ \begin{aligned} y &= (a+x)(b-y) =(b+x)(a-y) \\ &= \frac{b^2-x^2}{b-x}\cdot \frac{a^2-y^2}{a+y} =\frac{yx}{(b-x)(a+y)} \\ &=\frac {yx}y=x\ . \end{aligned} $$ $\square$
