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

[Emeric Deutsch]

What about extracting n-th root on both sides of (x+1)^n = x^2n ?
Emeric  

On Thu, 27 May 2004, cino hilliard wrote:

> Hi,
> While driving home from MicroCenter, I pondered on the golden ratio or the 
> solution to the
> proportion: the lesser is to the greater as the greater is to the whole. 
> Then me thought what
> if I say the square of the lesser is to the square of the greater as the 
> square or the greater is to
> the square of the whole? This was a little heavy while driving so I waited 
> until I got home.
> 
> There  I jotted out   a^2/x^2 = x^2/(a+x)^2 and lo and behold for a =1 we 
> get 1/x^2 = x^2/
> (x+1)^2 or  (x+1)^2 - x^4 = 0. Lo and behold we get the roots 1/2 - 
> sqrt(5)/2 and 1/2 + sqrt
> (5)/2. Curious I tried cubes, fourth powers etc and still got the same two 
> roots. So I concluded this
> was generally true:
> 
>                                 (x+1)^n - x^2n = 0 has real roots
>                                 x1 = 1/2 - sqrt(5)/2
>                                 x2 = 1/2 + sqrt(5)/2
> for all n.
> 
> Proof anyone?
> 
> Maple    solve(((x+1)^n - x^(n+n) = 0,x);
> 
>                                  1/2 - sqrt(5)/2 ,1/2 + sqrt(5)/2
> 
> gives the result. Can this be considered a proof?
> 
> There are other interesting curios for (x+a)^n - x^2n = 0 has real roots 
> when a = 2,3,4 etc but different powers give different roots.
> 
> Have fun,
> 
> Cino
> 
> 
>