If $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$, show that $x+y=0$ 
For $\{x,y\}\subset  \Bbb R$, $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1.$
Prove that $x+y=0.$

Problem presented in a book, as being from Norway Math Olympiad 1985. No answer was presented. My developments are not leading to a productive direction. Sorry if this is a duplicate. Hints and answers are welcomed.

Multiplying by $(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})$ we get 
$$(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})=1$$
and thus $$(x+\sqrt{x^2 +1})(y+\sqrt{y^2 +1})=(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})$$
hence 
$$x(\sqrt{x^2 +1} +\sqrt{y^2 +1} )=-y(\sqrt{x^2 +1} +\sqrt{y^2 +1} )$$
therefore $$x=-y$$
