If $$x = \frac{\sqrt{5}-2}{\sqrt{5}+2}$$, then $$x^4 + x^{-4}$$ is

Given : $$x = \frac{\sqrt{5}-2}{\sqrt{5}+2}$$

=> $$x = \frac{\sqrt{5}-2}{\sqrt{5}+2}\times(\frac{\sqrt5-2}{\sqrt5-2})$$

=> $$x=\frac{(\sqrt5-2)^2}{5-4}$$

=> $$x=9-4\sqrt5$$ ------------(i)

Similarly, $$\frac{1}{x}=9+4\sqrt5$$ -------------(ii)

To find : $$x^4 + x^{-4}=x^4+\frac{1}{x^4}$$

= $$(x^2+\frac{1}{x^2})^2-2$$

= $$[(x+\frac{1}{x})^2-2]^2-2$$

Substituting values from equations (i) and (ii), we get :

= $$[(9-4\sqrt5+9+4\sqrt5)^2-2]^2-2$$

= $$[324-2]^2-2$$

= $$103684-2=103682$$, which is an integer.

=> Ans - (C)