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. 

$$
p = x + \sqrt{x^2+3} \Rightarrow \sqrt{x^2+3} = p-x \\
p^2 = x^2 + x^2+3 + 2x \sqrt{x^2+3} = 2x^2+3 + 2x(p-x) = 2xp+3 \\
x = \frac {p^2-3}{2p}
$$
Analogously, if $y + \sqrt{y^2+3} = q$, then
$$
y = \frac {q^2-3}{2q}
$$
Initial expressions becomes $pq = 3$. Now, if calculate $x+y$ you'll get
$$
x+y = \frac {p^2-3}{2p} + \frac {q^2-3}{2q} = \frac 12 \left (\frac {p^2q-3q+q^2p-3p}{pq} \right ) = \frac {p+q}2 \left( pq-3\right) = 0
$$
