First degree is equal to second degree?

Since $x^{2} + \dfrac{1}{x^{2}} = \left(x + \dfrac{1}{x}\right)^{2} - 2$ , if we let $x + \dfrac{1}{x} = y$ the given equation becomes

$y = y^{2} - 2 \Longrightarrow y^{2} - y - 2 = (y - 2)(y + 1) = 0$ ,

so either $y = 2$ or $y = -1$ . But as $x \gt 0$ we must have $y \gt 0$ , and so

$y = x + \dfrac{1}{x} = 2 \Longrightarrow x^{2} + 1 = 2x \Longrightarrow x^{2} - 2x + 1 = (x - 1)^{2} = 0 \Longrightarrow x = 1$ .

Thus $x^{1024} + \dfrac{1}{x^{1024}} = 1 + \dfrac{1}{1} = \boxed{2}$ .

$x+ \frac{1}{x} = x² + \frac{1}{x²}$

Multiply both side by $x²$

$x³ + x = x^4 + 1$

$x^4- x³ - x +1 =0$

If $x=1$ , the equation is complete. Then, we have $x=1$ .

Hence,

$x^{1024} + \frac{1}{x^{1024}} = 1 + 1$

$= \boxed{\boxed{2}}$