compute sums of x,y given a condition 
Problem: given that $\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=p$, then compute $x+y$

try: i tryed to solve by this way
$$\begin{align}
\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)&=p\\
1&=p\left(x-\sqrt{x^2+1}\right)\left(y-\sqrt{y^2+1}\right)\\
p&=p^2\left(x-\sqrt{x^2+1}\right)\left(y-\sqrt{y^2+1}\right)\\
\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)&=p^2\left(x-\sqrt{x^2+1}\right)\left(y-\sqrt{y^2+1}\right)
\end{align}$$
however this only given
$$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=xy+x\sqrt{y^2+1}+y\sqrt{x^2+1}+\sqrt{(x^2+1)(y^2+1)}$$
$$\left(x-\sqrt{x^2+1}\right)\left(y-\sqrt{y^2+1}\right)=xy-x\sqrt{y^2+1}-y\sqrt{x^2+1}+\sqrt{(x^2+1)(y^2+1)}$$
wich not help, i tryed a lot of think but i alway end like walking in circles, how i solve this problem

It is not possible. We have that $f(x):\mathbb{R}\to\mathbb{R}^+$ given by:
$$f(x)=x+\sqrt{x^2+1}$$
is bijective. So, assume that $p=f(2)\cdot f(3)=6+5 \sqrt{2}+3 \sqrt{5}+2 \sqrt{10}$.
Obviously $x=2,y=3$ is a solution of $f(x)\cdot f(y)=p$, giving $x+y=5$.
But another solution exists, given by:
$$x=y=f^{-1}\left(\sqrt{6+5 \sqrt{2}+3 \sqrt{5}+2 \sqrt{10}}\right).$$
Since $f^{-1}(x)=\frac{x^2-1}{2x}$, the solution:
$$ x = y = \sqrt{\frac{5}{2} \left(1+\sqrt{2}\right)} $$
gives:
$$ x+y = \sqrt{10(1+\sqrt{2})} \neq 5. $$
