The Index Is $n$

I've reported this problem.

But, I will know it as $\displaystyle x + \frac { 1 } { x } = 2$ .

Then, multiplying $x$ to each number.

And, we get $x ^ { 2 } + 1 = 2x$ .

So, $x ^ { 2 } - 2x + 1 = 0 \Rightarrow ( x - 1 ) ^ { 2 } = 0 \Rightarrow x - 1 = 0 \Rightarrow x = 1$ .

Then, $\displaystyle 1 ^ { n } + ( \frac { 1 } { 1 } ) ^ { n } = 1 ^ { n } + 1 ^ { n }$ .

If the base is one, then it is always one despite the exponents are any number excepting $\infty$ .

But, $\infty$ is not in the whole real numbers, so $1 + 1 = \boxed { 2 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! }$

• We find the value of $x$ first

$\dfrac{x+1}{x} = 2 \quad \implies \quad x+1 = 2x \quad [x \ne 0] \quad \implies \quad x = 1$

• The value to find is

$x^n + \dfrac{1}{x}^n \implies 1^n + 1^n = \boxed{2}$

Since $x$ is clearly positive, we can apply AM-GM:

$\begin{aligned} \dfrac{x + \dfrac{1}{x}}{2} &\ge \sqrt{x \cdot \dfrac{1}{x}} \\ \\ \dfrac{x + \dfrac{1}{x}}{2} &\ge 1 \\ \\ x + \dfrac{1}{x} &\ge 2 \end{aligned}$

with equality iff $x = \dfrac{1}{x}$ . Thus $x^2 = 1$ , so $x = 1$ (as $x$ can't be negative.) $\$ Our answer then is $x^n + \left( \dfrac{1}{x} \right)^n = 1^n + 1^n = \boxed{2}$