Rational Number: Proving $x+\dfrac{1}{x}$ is Rational

[anemone]

Let $x$ be a non-zero number such that $x^4+\dfrac{1}{x^4}$ and $x^5+\dfrac{1}{x^5}$ are both rational numbers. Prove that $x+\dfrac{1}{x}$ is a rational number.

 

[Opalg]

Spoiler

For each integer $n$, let $R_n = x^n + \dfrac1{x^n}$. If $m>n$ then $$R_mR_n = \left( x^m + \dfrac1{x^m}\right)\left( x^n + \dfrac1{x^n}\right) = x^{m+n} + x^{m-n} + \dfrac1{x^{m-n}} + \dfrac1{x^{m+n}} = R_{m+n} + R_{m-n}.$$ In the case when $m=n$ that becomes $R_{2n} = x^{2n} + \dfrac1{x^{2n}} = \left( x^n + \dfrac1{x^n}\right)^2 - 2 = R_n^2 - 2.$

Given that $R_4$ and $R_5$ are rational, it follows that $R_8$ and $R_{10}$ are rational.

Next, $R_8R_2 = R_{10} + R_6$ and $R_4R_2 = R_6 + R_2$. So $R_8R_2 = R_{10} + (R_4-1)R_2$ and therefore $R_2 = \dfrac{R_{10}}{R_8 - R_4 + 1}$. Hence $R_2$ is rational, and so is $R_6 = (R_4-1)R_2$.

Finally, $R_5R_1 = R_6 + R_4$, so that $R_1 = \dfrac{R_6+R_4}{R_5}$, which is rational.