# If (x+sqrt(x^2+1))(y+sqrt(y^2+1))=1 what is x+y?

Sep 3, 2016

$x + y = 0$

#### Explanation:

Let $y = - x$

Then:

$\left(x + \sqrt{{x}^{2} + 1}\right) \left(y + \sqrt{{y}^{2} + 1}\right) = \left(\sqrt{{x}^{2} + 1} + x\right) \left(\sqrt{{x}^{2} + 1} - x\right)$

$\textcolor{w h i t e}{\left(x + \sqrt{{x}^{2} + 1}\right) \left(y + \sqrt{{y}^{2} + 1}\right)} = \left({x}^{2} + 1\right) - {x}^{2}$

$\textcolor{w h i t e}{\left(x + \sqrt{{x}^{2} + 1}\right) \left(y + \sqrt{{y}^{2} + 1}\right)} = 1$

So $x + y = 0$ always results in $\left(x + \sqrt{{x}^{2} + 1}\right) \left(y + \sqrt{{y}^{2} + 1}\right) = 1$

To see that this is the only possible value of $x + y$, note that the function $f \left(x\right) = x + \sqrt{{x}^{2} + 1}$ is strictly monotonic increasing.

So if ${x}_{1} + \sqrt{{x}_{1}^{2} + 1} = {x}_{2} + \sqrt{{x}_{2}^{2} + 1}$ then ${x}_{1} = {x}_{2}$

Sep 3, 2016

$x + y = 0$

#### Explanation:

If $x + \sqrt{{x}^{2} + 1} = \delta$ then $y + \sqrt{{y}^{2} + 1} = \frac{1}{\delta}$

Solving

${\left(x - \delta\right)}^{2} = {x}^{2} + 1$ and

${\left(y - \frac{1}{\delta}\right)}^{2} = {y}^{2} + 1$

we obtain

$x = \frac{{\delta}^{2} - 1}{2 \delta}$ and
$y = - \frac{{\delta}^{2} - 1}{2 \delta}$

then

$x + y = 0$