<s>[INST] 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.
 [/INST] Since
$$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1  / \cdot (x-\sqrt{x^2+1})$$
we get 
$$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x$$ so
$$\sqrt{x^2+1}-\sqrt{y^2+1}=x+y\;/^2$$
and thus $$-\sqrt{y^2+1}\cdot \sqrt{x^2+1}+1= xy$$
so $$x^2y^2 +x^2+y^2+1=x^2y^2-2xy+1$$
and finaly $(x+y)^2=0$ so $x+y=0$
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