For $(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$, compute $x+y$ . I am trying to find $x+y$ given that
$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3.$$
It is the radicals in $\sqrt{x^2 +3}, \sqrt{y^2+3}$ that is bugging me. I tried to expand the left hand side 
$$ xy + y\sqrt{x^2+3} + x\sqrt{y^2 +3} + \sqrt{(x^2+3)(y^2+3)}$$
and see if a term $x+y$ comes out, but it looks hopeless at this point. 

Hint: Calculate $(-t+\sqrt{t^2+3})(t+\sqrt{t^2+3})$.
Details: The product in the hint is $3$.  So for every $y$, by putting $x=-y$ we get a solution of the equation 
$$(x+\sqrt{x^2+3})(y+\sqrt{y^2+3}=3)=3\tag{$1$}$$
with $x+y=0$. We now show these are the only solutions,
One of the terms in $(1)$ is $\le \sqrt{3}$, say the first one. Then the second one is $\ge \sqrt{3}$.  But for positive $t$, the function $t+\sqrt{t^2+3}$ is increasing, so there is a unique value of $y$ corresponding to $x$.  
