<s>[INST] Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$ Let
$$\left({x+\sqrt{x^2+1}}\right)\left({y+\sqrt{y^2+1}}\right)=1$$
Prove that $x+y=0$.


This is my solution:
Let
$$a=x+\sqrt{x^2+1}$$
and
$$b=y+\sqrt{y^2+1}$$
Then $x=\dfrac{a^2-1}{2a}$ and $y=\dfrac{b^2-1}{2b}$. Now $ab=1\implies b=\dfrac1a$. Then I replaced $x$ and $y$:
$$x+y=\dfrac{a^2-1}{2a}+\dfrac{b^2-1}{2b}=\dfrac{a^2-1}{2a}+\dfrac{\dfrac{1}{a^2}-1}{\dfrac{2}{a}}=0$$
This solution is absolutely different from solution in my book. Is my solution mathematically correct? Did I assumed something that may not be true?
 [/INST] Given the symmetry of the problem, wlog $x\ge y$. $x=y=0$ is a solution, from now on let us assume that $x>0$.
Let $f$ be the real-valued map $w\mapsto w+\sqrt{w^2+1}$. If $w>0$ then $f(w)>1$, which is enough to prove that $x>0>y$. Even better (without deriving it), $f$ is strictly increasing on the positive reals.
Define $z=-y>0$. Then $f(x)f(y)=1$ is equivalent to (just rationalize it) 
$$f(x)=f(z),$$
which implies that $x=z$. This is equivalent to $x+y=0$.
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