Question

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

Answer 1

`x+y = 0`

Explanation:

Let `y = -x`

Then:

`(x+sqrt(x^2+1))(y+sqrt(y^2+1)) = (sqrt(x^2+1)+x)(sqrt(x^2+1)-x)`

`color(white)((x+sqrt(x^2+1))(y+sqrt(y^2+1))) = (x^2+1)-x^2`

`color(white)((x+sqrt(x^2+1))(y+sqrt(y^2+1))) = 1`

So `x+y = 0` always results in `(x+sqrt(x^2+1))(y+sqrt(y^2+1)) = 1`

To see that this is the only possible value of `x+y`, note that the function `f(x) = 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`

Answer 2

`x+y=0`

Explanation:

If `x+sqrt(x^2+1) = delta` then `y+sqrt(y^2+1) = 1/delta`

Solving

`(x-delta)^2=x^2+1` and

`(y-1/delta)^2=y^2+1`

we obtain

`x = (delta^2-1)/(2delta)` and
`y =- (delta^2-1)/(2delta)`

then

`x+y=0`