If $x,y,z$ are real numbers satisfying relations
\[x+y+z = 1 \quad \textrm{and} \quad \arctan x + \arctan y + \arctan z = \frac{\pi}{4},\]
prove that $x^{2n+1} + y^{2n+1} + z^{2n+1} = 1$ holds for all positive integers $n$.
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The trigonometric relation is equivalent to
\[\frac{x+y+z-xyz}{1-xy-xz-yz}=1\]
or $xyz-xy-xz-yz-x-y-z+1=0$. But since $x+y+z=1$,
$0=xyz-xy-xz-yx+x+y+z-1=(x-1)(y-1)(z-1)$.
Hence one of $x, y, z$ is $1$. WLOG suppose $x=1$. Now $y=-z$ and hence $x^{2n+1}+y^{2n+1}+z^{2n+1}=1+y^{2n+1}-y^{2n+1}=1$.
Quote:
The trigonometric relation is equivalent to
\frac{x+y+z-xyz}{1-xy-xz-yz}=1
I must be missing something trivial... Can you show, how do you get this?