Check for Substitution

$\begin{aligned} X & = \frac {2x}{1-x^2} + \frac {2y}{1-y^2} + \frac {2z}{1-z^2} \\ & = \frac {\displaystyle \sum_{cyc} 2x(1-y^2)(1-z^2)}{(1-x^2)(1-y^2)(1-z^2)} \\ & = \frac {2 \sum \left(x-xy^2 -z^2x+xy^2z^2\right)}{(1-x^2)(1-y^2)(1-z^2)} \\ & = \frac {2\left({\color{#3D99F6}x+y+z}-xy^2-yz^2-zx^2 -x^2y-y^2z-z^2x+xy^2z^2+x^2yz^2+x^2y^2z \right)}{(1-x^2)(1-y^2)(1-z^2)} \\ & = \frac {2\left({\color{#3D99F6}xyz}-xy(x+y)-yz(y+z)-zx(z+x)+xyz(xy+yz+zx) \right)}{(1-x^2)(1-y^2)(1-z^2)} \\ & = \frac {2\left(xyz-xy(x+y+{\color{#D61F06}z})-yz({\color{#D61F06}x}+y+z)-zx(x+{\color{#D61F06}y}+z) {\color{#D61F06}+3xyz}+xyz(xy+yz+zx) \right)}{(1-x^2)(1-y^2)(1-z^2)} \\ & = \frac {2\left(4xyz - {\color{#3D99F6}(x+y+z)}(xy+yz+zx) +xyz(xy+yz+zx) \right)}{(1-x^2)(1-y^2)(1-z^2)} \\ & = \frac {2\left(4xyz - {\color{#3D99F6}xyz}(xy+yz+zx) +xyz(xy+yz+zx) \right)}{(1-x^2)(1-y^2)(1-z^2)} \\ & = \frac {8xyz}{(1-x^2)(1-y^2)(1-z^2)} \end{aligned}$

$\implies n = \boxed{8}$

Proposition: Let $a,b,c \in \mathbb{R}$ . Then $\tan a+\tan b+\tan c=\tan a\tan b\tan c$ if and only if $a+b+c=\pi k$ for some $k \in \mathbb{Z}$ .

Proof:

• Suppose that $a+b+c=\pi k$ for some $k \in \mathbb{Z}$ . Then $\tan(a+b)=\tan(\pi k-c)$ . Since $\tan(x)$ is an odd function and has period $\pi$ , then $\tan(a+b)=-\tan(c)$ . Using the formula for the tangent of the sum of two angles we have $\dfrac{\tan a+\tan b}{1-\tan a\tan b}=-\tan c$ . Finally we rearrange the equation: $\tan a+\tan b=-\tan c+\tan a\tan b\tan c$ , so $\tan a+\tan b+\tan c=\tan a\tan b\tan c$ .

• Suppose that $\tan a+\tan b+\tan c=\tan a\tan b\tan c$ . Then $\tan a+\tan b=-\tan c+\tan a\tan b\tan c$ . We can factor the RHS to obtain $\tan a+\tan b=-\tan c(1-\tan a\tan b)$ , so $\dfrac{\tan a+\tan b}{1-\tan a\tan b}=-\tan c$ . Using the formula that we used before and the fact that $\tan x$ is odd we have $\tan(a+b)=\tan(-c)$ , so $a+b+\pi m=-c+\pi n$ for all $m,n \in \mathbb{Z}$ . Finally, $a+b+c=\pi(n-m)$ , and $n-m \in \mathbb{Z}$ .

Let $x=\tan a$ , $y=\tan b$ and $z=\tan c$ . Let the expression be $X$ . Then $X=\dfrac{2\tan a}{1-\tan^2 a}+\dfrac{2\tan b}{1-\tan^2 b}+\dfrac{2\tan c}{1-\tan^2 c}=\tan(2a)+\tan(2b)+\tan(2c)$ . Since $x+y+z=xyz$ , we have $a+b+c=\pi k$ for some $k \in \mathbb{Z}$ and $2a+2b+2c=2\pi k$ . Then $X=\tan(2a)\tan(2b)\tan(2c)=\left(\dfrac{2\tan a}{1-\tan^2 a}\right) \left(\dfrac{2\tan b}{1-\tan^2 b}\right) \left(\dfrac{2\tan c}{1-\tan^2 c}\right)=\dfrac{2xyz}{(1-x^2)(1-y^2)(1-z^2)}$ , so $n=\boxed{8}$ .