find x+y if $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=42 $? Let $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=42 $. 
How find $x+y$? I tried a number of ways.
 A: Let's call $\alpha$ the value of $x+\sqrt{x^2+1}$ and $\beta$ the value of $y+\sqrt{y^2+1}$
Then we know that $x=\dfrac{\alpha^2-1}{2\alpha}$ and $y=\dfrac{\beta^2-1}{2\beta}$
We know also that $\alpha\beta=42$
Hence $x+y=\dfrac{\alpha^2-1}{2\alpha}+\dfrac{\beta^2-1}{2\beta}=\dfrac{(\alpha\beta-1)(\alpha+\beta)}{2\alpha\beta}$
$x+y=\dfrac{41}{84}\times(\alpha+\beta)=\dfrac{41}{84}\times(\dfrac{42}{\beta}+\beta)$
$x+y$ can thus take many different values
A: There isn't just one possible value for $x + y$, which we can see as follows. Let $a = x + \sqrt{x^2 + 1}$ and $b = y + \sqrt{y^2 + 1}$. Then ${1 \over a} = -x + \sqrt{x^2 + 1}$ and ${1 \over b} = -y + \sqrt{y^2 + 1}$. As a result we have
$$x + y = {1 \over 2}\bigg(a - {1 \over a}\bigg) + {1 \over 2}\bigg(b - {1 \over b}\bigg)$$
$$= {1 \over 2}\bigg({a^2 - 1 \over a}\bigg) + {1 \over 2}\bigg({b^2 - 1 \over b}\bigg)$$
$$= {1 \over 2}{a^2b + ab^2 - a - b \over ab}$$
$$= {1 \over 2}{(ab - 1)(a + b) \over ab}$$
The condition given is that $ab = 42$ so this is the same as
$$ {41 \over 84} (a + b)$$
Hence if $x+ y$ is uniquely determined, $a + b$ must also be uniquely determined. But all we are given is that $ab = 42$, and there are many pairs $(a,b)$ which satisfy $ab = 42$, each of which will have a different value for $a + b$. Since $z \rightarrow z + \sqrt{z^2 + 1}$ is monotone with range $[1,\infty)$, each such $(a,b)$  with $a,b \geq 1$ will correspond to some value of $x$ and $y$. 
Hence there is not just one answer to your question.
A: If $x=0$ the first parenthesis is $1$, so the second has to be $42$, WA says that means $y=\frac{1763}{84}\approx 20{,}988$.
If $x=1$ the first parenthesis is $1+\sqrt2$, so the second has to be $\frac{42}{1+\sqrt2}$, WA says that means $y=\frac{1}{84}(1763\sqrt2-1765)\approx 8{,}6697$.
$x+y$ doesn't sum to the same in these two cases, so the problem isn't well-defined.
