$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$. Find $(x+y)$.

Question

We know that $\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$. Find the expression $(x+y)$.

My work so far:

$$\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1$$ $$\color{red}{\left(x-\sqrt{x^2+1}\right)\cdot}\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=\color{red}{\left(x-\sqrt{x^2+1}\right)\cdot}1$$ $$-\left(y+\sqrt{y^2+1}\right)=\left(x-\sqrt{x^2+1}\right)$$ $$-\color{red}{\left(y-\sqrt{y^2+1}\right)\cdot}\left(y+\sqrt{y^2+1}\right)=\color{red}{\left(y-\sqrt{y^2+1}\right)\cdot}\left(x-\sqrt{x^2+1}\right)$$ $$1=\left(x-\sqrt{x^2+1}\right)\left(y-\sqrt{y^2+1}\right)$$ I need help here.

Answer 1

Notice that we have $$-\left(y+\sqrt{y^2+1}\right)=\left(x-\sqrt{x^2+1}\right)$$ From your third line. Now $$x+y=\sqrt{x^2+1}-\sqrt{y^2+1}\tag{1}$$ In the same fashion, we can get$$x+y=\sqrt{y^2+1}-\sqrt{x^2+1}\tag{2}$$ Now adding $\text{(1)}$ and $\text{(2)}$ together gives $x+y=0$. We are done.

Answer 2

$t\mapsto t+\sqrt{t^2+1}$ is a strictly increasing function, hence for each $x$ there is at most one $y$ that makes the equation true. On the other hand, $$(x+\sqrt{x^2+1})(-x+\sqrt{x^2+1})=(x^2+1)-x^2=1$$ suggests that $y=-x$ is a valid such choice. We conclude that $x+y=0$.