Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$

Question

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?

Answer 1

Note $$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\tag{1}$$ $$x+\sqrt{x^2+1}=\sqrt{y^2+1}-y\tag{2}$$ $(1)+(2)$ $$\Longrightarrow x+y=-(x+y)$$ $$\Longrightarrow x+y=0$$

Answer 2

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 $x+y=0$.

Answer 3

Hint: Let $x=\sinh a$ and $y=\sinh b$. Then, using the fact that $\cosh^2u-\sinh^2u=1$ and $\sinh u+\cosh u=e^u$, we arrive at $e^{a+b}=1\iff a+b=0$, assuming a and b are reals.
Since $\sinh u$, just like $\sin u$, is an odd function, the proof is complete.