2022/045) Find x given $(2+\sqrt{3})^x + (2-\sqrt{3})^x = 4$

 we see that $(2+\sqrt{3})(2-\sqrt{3}) = 4 - 1 = 1$

so $\frac{1}{(2+\sqrt{3}} = 2 - \sqrt{3}$

Let $(2+\sqrt{3})^x = y$ so $(2-\sqrt{3})^x = \frac{1}{y}$

so $y +\frac{1}{y} = 4$

or $y^2 - 4y + 1 = 9$

ot $y = \frac{4 \pm \sqrt{16-4}}{2}$

or $y= 2 \pm \sqrt{3}$

If we take $y = 2 + \sqrt{3}$ then $(2+\sqrt{3})^x = 2 + \sqrt{3}$ or $x= 1$

 Taking $y = 2 - \sqrt{3}$ then $(2+\sqrt{3})^x = 2 - \sqrt{3}$ or $x= -1$

so $x \in \{ 1, -1\}$