If $x = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}}$, then using the properties of proportion prove that $x^2 - 4ax + 1 = 0$
Given: $\;$ $x = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}}$
i.e. $\;$ $\dfrac{x}{1} = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}}$
We have by componendo-dividendo,
$\dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1} + \sqrt{2a + 1} - \sqrt{2a - 1}}{\sqrt{2a + 1} + \sqrt{2a - 1} - \sqrt{2a + 1} + \sqrt{2a - 1}}$
i.e. $\;$ $\dfrac{x + 1}{x - 1} = \dfrac{2 \sqrt{2a + 1}}{2 \sqrt{2a - 1}}$
i.e. $\;$ $\dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 1}}{\sqrt{2a - 1}}$ $\;\;\; \cdots \; (1)$
Squaring both sides of equation $(1)$ gives,
$\dfrac{\left(x + 1\right)^2}{\left(x - 1\right)^2} = \dfrac{2a + 1}{2a - 1}$
i.e. $\;$ $\dfrac{x^2 + 2x + 1}{x^2 - 2x + 1} = \dfrac{2a + 1}{2a - 1}$
We have by componendo-dividendo,
$\dfrac{x^2 + 2x + 1 + x^2 - 2x + 1}{x^2 + 2x + 1 - x^2 + 2x - 1} = \dfrac{2a + 1 + 2a -1}{2a + 1 - 2a + 1}$
i.e. $\;$ $\dfrac{2x^2 + 2}{4x} = \dfrac{4a}{2}$
i.e. $\;$ $\dfrac{x^2 + 1}{2x} = 2a$
i.e. $\;$ $x^2 + 1 = 4ax$
or, $\;$ $x^2 - 4ax + 1 = 0$
Hence proved.