Roots of (x+1)^n - x^2n = 0

[jens at voss-ahrensburg.de]

> > What about extracting n-th root on both sides of (x+1)^n = x^2n ?
> > Emeric
> Indeed, x+1 = x^2 or x^2 - x - 1 = 0 has the the 2 real roots. I
should 
> have figured that out.

Alternatively,

x^(2n+1) - (x+1)^(n+1) = x^(2n) * (x^2-x-1) + [x^(2n) - (x+1)^n] *
(x+1).

Now use induction.

Regards,
Jens
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