Manipulate the 2

Solution 1: Finding Value for $x$

$\begin{aligned} x + \frac{1}{x} &= 2 \quad \color{#3D99F6}{\text{Given}} \\ \\ \frac{x^2 + 1}{x} &= 2 \quad \color{#3D99F6}{\text{Square the expression}} \\ \\ x^2 + 1 &= 2x \quad \color{#3D99F6}{\text{Cross multiply}} \\ \\ x^2 - 2x + 1 &= 0 \quad \color{#3D99F6}{\text{Bring 2x to LHS}} \\ \\ (x-1)^2 &= 0 \quad \color{#3D99F6}{\text{Factorise}} \\ \\ x &= 1 \quad \color{#3D99F6}{\text{Solve for x}} \\ \\ \\ (1)^{2n} + \frac{1}{(1)^{2n}} &\quad \color{#3D99F6}{\text{Substitute back into the equation}} \\ \\ = \boxed{2} &\quad \color{#3D99F6}{[\because(1)^{2n} = 1 ]} \end{aligned}$

Solution 2: Manipulation

$\begin{aligned} x + \frac{1}{x} &= 2 \quad \color{#D61F06}{\text{Given}} \\ \\ \bigg(x + \frac{1}{x} \bigg)^{2} &= (2)^{2} \quad \color{#D61F06}{\text{Square the expression}} \\ \\ x^2 + \frac{1}{x^2} + \bigg(2 \cdot x \cdot \frac{1}{x}\bigg) &= 4 \quad \color{#D61F06}{\text{Expand and Simplify}} \\ \\ x^2 + \frac{1}{x^2} &= 2 \quad \color{#D61F06}{\text{For } n=1} \\ &. \\ &. \\ &. \\ &. \\ &\color{#D61F06}{\text{Repeat for } n \text{ times}} \\ &\color{#D61F06}{\text{Regardless of }n, \text{ answer is always 2}}\end{aligned}$

$x + \dfrac 1x = 2 \implies x^2 -2x + 1 = 0 \implies (x-1)^2 = 0 \implies x = 1\implies x^n + \dfrac 1{x^n} = \boxed 2$ for any $n \in \mathbb Z$ .

We know that $x+\dfrac{1}{x}\geq 2\sqrt {x\times \dfrac{1}{x}}=2$ . The minimum is attained for $x=1$ . So, for any natural number $p, x^p+\dfrac{1}{x^p}=1+1=2$ . Hence, for $p=2n, x^{2n}+\dfrac{1}{x^{2n}}=\boxed 2$ .