rationalskepticism.org

Going off on a completely different tack:

Consider the quadratic equation, x^2 - x - k = 0

Solving that equation yields the solutions: x1=(1+Sqrt(4k+1))/2 and x2= (1-Sqrt(4k+1))/2

Also consider that: x^2 = x + k.

Now consider the n=2 case from the original expression which can be rewritten as (x1^2 - x2^2)/Sqrt(4k+1)
The x1^2 term can be substituted with (x1+ k) and the x2^2 term can be substituted with (x2 + k)

Therefore the n=2 case of the expression can be written as
(x1 + k - (x2 + k))/Sqrt(4k+1)
= (x1 - x2)/Sqrt(4k+1)
= (1 + Sqrt(4k+1) - (1 - Sqrt(4k+1))/(2*Sqrt(4k+1))
= (2*Sqrt(4k+1))/(2*Sqrt(4k+1)
= 1

Now consider the n=3 case of the original expression:
x^3 = x*x*2 = x(x+k) = x^2 + xk = (x+k) + xk

Therefore the n=3 case of the expression can be written as
(x1 + k + (x1)k - (x2 + k + (x2)k)/Sqrt(4k+1)
= (x1 - x2 + (x1)k - (x2)k)/Sqrt(4k+1)
= ((x1-x2) + k(x1-x2))/Sqrt(4k+1)
= ((x1-x2)(k+1)/Sqrt(4k+1)
= k+1

Carrying on to the n=4 case:
x^4 = x*x^3 = x((x+k) + xk) = x^2 + kx + kx^2 = (x+k) + kx + kx + k^2 = x + 2kx + k^2

The n=4 case of the expression can be written as
(x1 + 2k(x1) + k^2 - (x2 + 2k(x2) + k^2))/Sqrt(4k+1)
= (x1 - x2 + 2k(x1 - x2))/Sqrt(4k+1)
= 2k+1

Does this boost the argument by induction?