1tan^{-1}x+tan^{-1}y+tan^{-1}z=tan^{-1}(\frac{x+y+z-xyz}{1-xy+yz+zx})=\frac{\pi}{4}\\\\\\\Rightarrow xyz=xy+yz+xz\\\\\\or,\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\\\\\\ multiplying\ both\\\\\\(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=1\\\\\\ x^2y+y^2x+x^2z+z^2x+y^2z+z^2y+2xyz=0\\\\\\(x+y)(xy+yz+xz+z^2)=0\\\\\\(x+y)(y+z)+(z+x)=0\\\\\\so,three\ casec\ arrises\\\\\\x=-y,\ z=1\\\\y=-z,\ x=1\\\\z=-x,\ y=1\\\\ hence,\ for\ all\ three\ cases\\\\\\x^{2n+1}+y^{2n+1}+z^{2n+1}=1
341you can reduce the steps a bit.
\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{4} \Rightarrow \tan (\tan^{-1} x + \tan^{-1} y + \tan^{-1} z ) = 1
\Rightarrow \frac{x+y+z-xyz}{1-\sum xy} = 1 \Rightarrow 1 - \sum x + \sum xy - xyz = 0
So, if x,y,z are the roots of the cubic x^3+ax^2+bx+c = 0, we get from the above relation that x=1 is a root of the equation.
i.e. at least one of x, y,z equals 1.
In any case WLOG x =1 and y =-z
and hence x^{2n+1} + y^{2n+1} + z^{2n+1} = 1 for all whole numbers n
