Rational or Irrational?

Suppose that there is an irrational number $'a'$ such that

$A=\sqrt[n]{a+ \sqrt{a^2-1}} + \sqrt[n]{a- \sqrt{a^2-1}}$ is rational

Let $\alpha = \sqrt[n]{a+ \sqrt{a^2-1}}$

Then $\frac{1}{\alpha} = \frac{1}{ \sqrt[n]{a+ \sqrt{a^2-1}}} = \sqrt[n]{a- \sqrt{a^2-1}}$ (By Rationalizing)

$\implies A= \alpha +\frac{1}{\alpha}$ is rational

We will prove that whenever $A$ is rational this implies $\alpha^n + \frac{1}{\alpha^n}$ is rational.

We check with the starting powers $2$ and $3$

$\alpha^2 + \frac{1}{\alpha^2} = (\alpha + \frac{1}{\alpha})^2 -2$ is rational

and $\alpha^3 + \frac{1}{\alpha^3}=(\alpha+ \frac{1}{\alpha})^3 - 3(\alpha+ \frac{1}{\alpha})$ is rational.

Using the identity $(\alpha^k+ \frac{1}{\alpha^k})=(\alpha^{k-1}+ \frac{1}{\alpha^{k-1}})(\alpha+ \frac{1}{\alpha})-(\alpha^{k-2}+ \frac{1}{\alpha^{k-2}})$

it follows by induction that $(\alpha^k+ \frac{1}{\alpha^k})$ is rational for all positive integers $k$ , hence $(\alpha^n+ \frac{1}{\alpha^n})$ is rational.

Thus $(a+\sqrt{a^2-1}+a-\sqrt{a^2-1})=2a$ is rational, which contradicts our assumption that $'a'$ is irrational.

We conclude that there is $NO$ irrational number $'a'$ such that

$A=\sqrt[n]{a+ \sqrt{a^2-1}} + \sqrt[n]{a- \sqrt{a^2-1}}$ is rational