The real numbers $x, y$ satisfy the equality: $$x\cdot\frac{4^x-2^y}{4^x+2^y} =y\cdot \frac{4^y-2^x}{4^y+2^x}$$ Prove that $|x| = |y|$
PS. I tried to work this way but ran out of ideas:( Maybe there is another slicker way? $$\frac{x}{y}\cdot \frac{4^x-2^y}{4^x+2^y} = \frac{y}{x}\cdot \frac{4^y-2^x}{4^y+2^x} \Rightarrow \left(\frac{x}{y}\cdot \frac{4^x-2^y}{4^x+2^y}\right)^{-1} = \left(\frac{y}{x}\cdot \frac{4^y-2^x}{4^y+2^x}\right)^{-1}$$
here is my method, which might not be the right way. First, the equality is equivalent to: $$ x(4^{x+y}-2^{x+y}+2^{3x}-2^{3y})=y(4^{x+y}-2^{x+y}-2^{3x}+2^{3y}) $$ which is equivalent to $$ (x-y)(4^{x+y}-2^{x+y})+(x+y)(2^{3x}-2^{3y})=0 $$ Next, notice that this equality holds if and only if $x-y=0$ or $x+y=0$. Because for example, if $x-y>0$ and $x+y>0$ then you will get positive + positive. if $x-y>0$ and $x+y<0$, you will get negative + negative.
