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

[cino hilliard]

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