If $x^2 + y^2 + \dfrac{1}{x^2} + \dfrac{1}{y^2} = 4$, then find the value of x² + y²

Given,

$\Rightarrow x^2 + y^2 + \dfrac{1}{x^2} + \dfrac{1}{y^2} = 4 \\[1em] \Rightarrow x^2 + y^2 + \dfrac{1}{x^2} + \dfrac{1}{y^2} - 4 = 0 \\[1em] \Rightarrow \Big(x^2 + \dfrac{1}{x^2} - 2\Big) + \Big(y^2 + \dfrac{1}{y^2} - 2\Big) = 0$

Using identity,

$a^2 + \dfrac{1}{a^2} - 2 = \Big(a - \dfrac{1}{a}\Big)^2$.

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

Hence,

$\Big(x - \dfrac{1}{x}\Big)^2 = 0 \text{ and } \Big(y - \dfrac{1}{y}\Big)^2 = 0$

$x - \dfrac{1}{x} = 0 \text{ and } y - \dfrac{1}{y} = 0$

Solving for x: $\Rightarrow x = \dfrac{1}{x} \\[1em] \Rightarrow x^2 = 1$

Solving for y: $\Rightarrow y = \dfrac{1}{y} \\[1em] \Rightarrow y^2 = 1$

Solving for x² + y²:

x² + y² = 1 + 1

x² + y² = 2

Hence, x² + y² = 2.