For $(x+\sqrt{x^2+3})(y+\sqrt{y^2+3})=3$, compute $x+y$ .

$$ \begin{align} x+\sqrt{x^2+3} &=\frac3{y+\sqrt{y^2+3}}\\ &=\frac3{y+\sqrt{y^2+3}}\frac{y-\sqrt{y^2+3}}{y-\sqrt{y^2+3}}\\ &=\frac3{-3}\left(y-\sqrt{y^2+3}\right)\\ &=-y+\sqrt{y^2+3}\tag{1} \end{align} $$ Similarly $$ y+\sqrt{y^2+3}=-x+\sqrt{x^2+3}\tag{2} $$ Add $(1)$ and $(2)$ and cancel the radicals.

I want to explain why this is actually a remarkable problem.

The question is to show that an algebraic curve, of degree that could potentially be as high as 8 (it is a product of two terms with square roots) defines a straight line -- at least for its real solutions.

The usual ways in which such a thing happens, and are represented fairly often in competition problems and book exercises, are:

• the equation represents the equality condition in an inequality of real numbers (for example, arithmetic mean equals geometric mean for some suitably constructed set of variables), or

• the equation defining the curve factorizes, with some of the factors having no real solutions (for example, our degree 8 curve might be $(x+y)^4$ multiplied by a degree $4$ polynomial that is positive for all real coordinates $(x,y)$).

Neither of these is the case here. Algebraic calculation, separating $x$ and $y$ to different sides of the equation, shows that we have $f(x) = f(-y)$ for the increasing function $f(x) = x + \sqrt{x^2+3}$, so that for real solutions $x = -y$. This is a nice argument, and it generalizes to $F(x) = x + \sqrt{x^2 + a^2}$ and the equation $F(x)F(y)=a^2$, but does it fit into some algebraic framework? The issue is what happens with the complex solutions, or when the equation is handled purely algebraically.

Introducing variables $X$ and $Y$ with $X^2 - x^2=3$ and $Y^2 - y^2=3$, there is a system of 3 quadratic equations in 4 unknowns. The surprise here is that purely algebraic calculations on the system in the polynomial ring $\mathbb{Z}[x,y,X,Y]$, lead to $3(x+y)=0$. This illustrates some of the complexities around Bezout's theorem; for affine and singular varieties you cannot naively determine the degree of an intersection by degree counting alone. The algebra is not hard in this case, but it is well worth going through the geometric, projective and scheme-theoretic descriptions of what is happening in this deceptively simple exercise.

I will edit the question to see if any of the 10 downvoters would like to change their minds in light of this information (assuming a non OP edit allows that).

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 $x$.