$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$ is true if and only if $x+y=0$ Prove that $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$ is true if and only if $x+y=0$  
I believe x and y could both be 0 as that satisfies the equations. Beyond that, I do not know how to prove this.

You have $$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1}) = 1$$
$$x+\sqrt{x^2+1} = \sqrt{y^2+1} - y$$
$$x+y = -\sqrt{x^2+1} + \sqrt{y^2+1}$$
$$(x+y)(\sqrt{x^2+1} + \sqrt{y^2+1}) = -x^2+y^2$$
$$(x+y)(\sqrt{x^2+1} + \sqrt{y^2+1} +x-y) = 0$$
Now, you get $x+y=0$ or $\sqrt{x^2+1} + \sqrt{y^2+1} +x-y = 0$. Can you go further?
