3 Suppose that for two real numbers $x$ and $y$ the following equality is true:
$$
\left(x+\sqrt{1+x^{2}}\right)\left(y+\sqrt{1+y^{2}}\right)=1 .
$$

Find (with proof) the value of $x+y$.
Solution 1 (Rationalize and Factor): Move the $y$ 's to the other side by dividing both sides by $y+$ $\sqrt{1+y^{2}}$ :
$$
x+\sqrt{1+x^{2}}=\frac{1}{y+\sqrt{1+y^{2}}} .
$$

Then get rid of the denominator on the right hand side by multiplying top and bottom of the fraction by $-y+\sqrt{1+y^{2}}$ :
$$
x+\sqrt{1+x^{2}}=\frac{\sqrt{\left.1+y^{2}\right)}-y}{\left(y+\sqrt{1+y^{2}}\right)\left(\sqrt{1+y^{2}}-y\right)}=\frac{\sqrt{\left.1+y^{2}\right)}-y}{\left(1+y^{2}\right)-y^{2}}=\sqrt{1+y^{2}}-y .
$$

Now move back $y$ to the left and the radical $\sqrt{1+x^{2}}$ to the right:
$$
x+y=\sqrt{1+y^{2}}-\sqrt{1+x^{2}} \text {. }
$$

Now, again "rationalize" the right hand side by multiplying it and dividing by $\sqrt{1+y^{2}}+\sqrt{1+x^{2}}$ :
$$
x+y=\frac{\left(\sqrt{1+y^{2}}-\sqrt{1+x^{2}}\right)\left(\sqrt{1+y^{2}}+\sqrt{1+x^{2}}\right)}{\sqrt{1+y^{2}}+\sqrt{1+x^{2}}}=\frac{\left(1+y^{2}\right)-\left(1+x^{2}\right)}{\sqrt{1+y^{2}}+\sqrt{1+x^{2}}}=\frac{y^{2}-x^{2}}{\sqrt{1+y^{2}}+\sqrt{1+x^{2}}} .
$$

Move the denominator over the left hand side and factor $y^{2}-x^{2}=(y-x)(y+x)$ on the right hand side:
$$
(x+y)\left(\sqrt{1+y^{2}}+\sqrt{1+x^{2}}\right)=(x+y)(x-y) .
$$

Finally, move the right hand side to the left and factor out $(x+y)$ :
$$
(x+y)\left(\sqrt{1+y^{2}}+\sqrt{1+x^{2}}-(x-y)\right)=0 \text {, or, equivalently, }(x+y)\left(\left(\sqrt{1+y^{2}}+y\right)+\left(\sqrt{1+x^{2}}-x\right)\right)=0 .
$$

Regardless of what $x$ and $y$ are, we always have $1+y^{2}>y^{2}$ and $1+x^{2}>x^{2}$, and hence $\sqrt{1+y^{2}}>|y|$ and $\sqrt{1+x^{2}}>|x|$, which implies $\sqrt{1+y^{2}}+y>0$ and $\sqrt{1+x^{2}}-x>0$. Thus, the second factor in the product above, $\left(\sqrt{1+y^{2}}+y\right)+\left(\sqrt{1+x^{2}}-x\right)$ is always positive. So the product can equal 0 only if $x+y=0$.

Solution 2: (Clever Manipulations) Suppose
$$
\left(x+\sqrt{1+x^{2}}\right)\left(y+\sqrt{1+y^{2}}\right)=1 .
$$

Multiply both sides by $-y+\sqrt{1+y^{2}}$ to get
$$
x+\sqrt{1+x^{2}}=-y+\sqrt{1+y^{2}} .
$$

Rearrange to get
$$
x+y=\sqrt{1+y^{2}}-\sqrt{1+x^{2}} .
$$

By a symmetrical argument, we also have
$$
x+y=\sqrt{1+x^{2}}-\sqrt{1+y^{2}} \text {. }
$$

Average (1) and (2) to conclude that $x+y=0$.

Solution 3: (Calculus) We note that $y=-x$ gives $\left(x+\sqrt{1+x^{2}}\right)\left(y+\sqrt{1+y^{2}}=1+x^{2}-x^{2}=1\right.$. Further the function $x+\sqrt{1+x^{2}}$ is monotone increasing (easy with calculus) so for a given $x$ there is at most one solution to $\left(x+\sqrt{1+x^{2}}\right)\left(y+\sqrt{1+y^{2}}=1\right.$, hence that solution always is $-x$.

Solution 4: (Trigonometry): The equality can't hold if $x$ and $y$ are of the same sign. If $x=0$ then $y=0$ and vice versa. So WLOG, assume $x>0 y<0$. There are angles $X$ and $Y$ in $(-\pi / 2, \pi / 2)$ such that $\cot X=x$ and $\cot Y=y$. Then $\sqrt{1+x^{2}}=\frac{1}{\sin X}$ and $\sqrt{1+y^{2}}=-\frac{1}{\sin Y}, x+\sqrt{1+x^{2}}=\frac{\cos X+1}{\sin X}=\cot X / 2$, $y+\sqrt{1+y^{2}}=\frac{\cos y-1}{\sin y}=\tan -Y / 2$, so $\tan (-Y / 2) \cot (X / 2)=1$, so $\tan X / 2=\tan -Y / 2$, so $X=-Y$.