<s>[INST] Let $x^3+\frac{1}{x^3}$ and $x^4+\frac{1}{x^4}$ are rational numbers. Show that $x+\frac{1}{x}$ is rational. 
$x^3+\dfrac{1}{x^3}$ and $x^4+\dfrac{1}{x^4}$ are rational numbers. Prove that $x+\dfrac{1}{x}$ is rational number.

My solution:

$x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-1\right)$
$x^4+\dfrac{1}{x^4}=x^4+2+\dfrac{1}{x^4}-2=\left(x^2+\dfrac{1}{x^2}\right)^2-1-1=\left(x^2+\dfrac{1}{x^2}-1\right)\left(x^2+\dfrac{1}{x^2}+1\right)-1$
Because $x^4+\dfrac{1}{x^4}$ is rational number so
$\left(x^2+\dfrac{1}{x^2}-1\right)$ is rational number too and
$x^3+\dfrac{1}{x^3}$ is rational number so
$\left(x+\dfrac{1}{x}\right)$ is rational number.

Am I wrong? Please check my solution, thank you.
 [/INST] With $\alpha= x+\frac{1}{x}$ we get
$$x^3 + \frac{1}{x^3} =\alpha^3 - 3 \alpha = u\\
x^4 + \frac{1}{x^4} = \alpha^4 - 4 \alpha^2 + 2 = v$$
The polynomials $t^3 - 3 t - u$ and $t^4 - 4 t^2 + 2 - v$ have a common root $\alpha$. If that were the only common root, then their gcd $t-\alpha$ would have rational coefficients. So assume that they have another common root $\beta \ne \alpha$. Now we get the system
$$(\alpha^3 - 3 \alpha) -(\beta^3 - 3 \beta) = 0\\
(\alpha^4 - 4 \alpha^2) -(\beta^4 - 4 \beta^2) = 0$$
that is
$$(\alpha - \beta) (\alpha^2 + \alpha \beta + \beta^2 - 3)=0 \\
(\alpha-\beta)(\alpha+ \beta) (\alpha^2 + \beta^2 - 4)=0$$
and since $\alpha- \beta \ne 0$ we get finitely many solutions
$\pm (\sqrt{3}, -\sqrt{3}), \pm(\sqrt{2-\sqrt{3}}, -\sqrt{2-\sqrt{3}}(2+\sqrt{3}) )$ and its conjugate.  For neither of these $\alpha$'s do we get both rational values for $x^3 + 1/x^3$, $x^4 + 1/x^4$.
</s>