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.

Question

$x^3+\dfrac{1}{x^3}$ 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.

Answer 1

As the commenters pointed out, a problem in your argument is that the implication "$(x^4+x^{-4})$ is rational" $\implies$ "$(x^2+x^{-2})$ is rational" is false. For example, it is possible that $x^2+x^{-2}=\sqrt5$ when, by your tricks, $x^4+x^{-4}=(x^2+x^{-2})^2-2=3$, a rational number. Of course, in this case $x^3+x^{-3}$ fails to be rational, but this underlines a problem in your logic.

One commenter suggested using a contrapositive argument. This may be a possibility, but a bit delicate to manage.

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My favorite approach (that generalizes to several variants of this question) is somewhat high browed in the sense that it uses properties of algebraic field extensions. Outlining it for the benefit of the readers familiar with that theory.

• As $x^3+1/x^3=q_3\in\Bbb{Q}$ the number $x$ is a zero of the polynomial $P_3(T)=T^6-q_3T^3+1$ with rational coefficients. Therefore $x$ is an algebraic number. Let $m(T)\in\Bbb{Q}[T]$ be its minimal polynomial.
• The other zeros of $P_3(T)$ are $x^{\pm1}\omega^j$, $\omega=e^{2\pi i/3}$ a primitive third root of unity, $j\in\{0,1,2\}$. Therefore the zeros of $m(T)$ are among these numbers.
• But, as $x^4+1/x^4=q_4\in\Bbb{Q}$, $x$ is also a zero of the polynomial $P_4(T)=T^8-q_4T^4+1$. The zeros of $P_4(T)$ are seen to be $x^{\pm1}i^j$, $j\in\{0,1,2,3\}$.
• The zeros of $m(T)$ are common zeros of both $P_3(T)$ and $P_4(T)$, so they can only be $x^{\pm1}$.
• But $$r(T):=(T-x)(T-\frac1x)=T^2-(x+\frac1x)T+1.$$ Either $x$ is rational, when the claim is trivial, or $m(T)=r(T)$, implying that the coefficients of $r(T)$ are rational. Here $x+\dfrac1x$ is the coefficient of the linear term of $r(T)$ and we are done in this case as well.

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Toying with Mathematica gives the following elementary method.

Write $u=x+\dfrac1x$. We are given that (see the linked thread) $$ A=x^3+\frac1{x^3}=u^3-3u $$ and $$ B=x^4+\frac1{x^4}=u^4-4u^2+2 $$ are rational. Hence so is the quantity $$ \frac{A^3+AB-3A}{B^2-1}=u. $$

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Explaining the reason why some formula like the one above writing $u$ in terms of $A$ and $B$ must exist. Let's treat $u$ as a variable, an element transcendental over $\Bbb{Q}$. Write $F=\Bbb{Q}(u)$ for the field of rational functions. It has subfields $K_1=\Bbb{Q}(A)$ and $K_2=\Bbb{Q}(B)$. We immediately see that the respective extension degrees are $[F:K_1]=3$ and $[F:K_2]=4$. By the tower law, the only possibility is then that the compositum $K_1K_2=\Bbb{Q}(A,B)$ must be all of $F$. Consequently $u$ can be written as a rational function (with rational coefficients) in $A$ and $B$. Observe that the representation is not unique, because $A$ and $B$ are algebraically dependent over $\Bbb{Q}$ given that they both reside in the same transcendence degree one extension $F$.

Answer 2

Let $a_k=x^k+\frac1{x^k}$， we have $\begin{cases}a_1a_4=a_5+a_3\\a_2a_3=a_5+a_1\\a_1a_3=a_4+a_2\end{cases}$

By writing $a_3=u,a_4=v$, we could rewrite the fomula as $\begin{cases}va_1+0a_2-a_5=u\\a_1-ua_2+a_5=0\\ua_1-a_2+0a_5=v\end{cases}$ Solving the linear equation of $a_1,a_2,a_5$ we could get they're all rational number

Answer 3

I constructed a possible elementary algebraic way:

Let,

$$\begin{cases}x+\frac 1x=v,\thinspace v \not \in \mathbb Q\\ x^2+\frac{1}{x^2}=u,\thinspace u\in\mathbb R\\ x^3+\frac{1}{x^3}=p,\thinspace p\in\mathbb Q\\ x^4+\frac{1}{x^4}=q,\thinspace q\in\mathbb Q\end{cases}$$

This implies,

$$pv=u+q\not\in \mathbb Q\implies u\not\in\mathbb Q\\$$

Then we have,

$$\begin{align}&uq=p^2-2+u\\ \implies &uq-u=p^2-2\\ \implies &u=\frac{p^2-2}{q-1}\in\mathbb Q\\ &\thinspace \text{A contradiction.}\end{align}$$

Answer 4

I wanted to write this answer with the thought that it might be cleaner than my previous answer.

We have,

$$\begin{align}&\begin{cases}x+\frac 1x=u,\thinspace u \in \mathbb R\\ x^2+\frac{1}{x^2}=v,\thinspace v\in\mathbb R\\ x^3+\frac{1}{x^3}=p,\thinspace p\in\mathbb Q\\ x^4+\frac{1}{x^4}=q,\thinspace q\in\mathbb Q\end{cases}\\ \implies &vq=p^2-2+v\\ \implies &vq-v=p^2-2\\ \implies &v=\frac{p^2-2}{q-1}\in\mathbb Q\\ \implies &up=q+v\\ \implies &u=\frac{q+v}{p}\in\mathbb Q.\end{align}$$

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Final answer:

$$u=\frac{p^2q-p^2-q+2}{p}\in\mathbb Q.$$

Answer 5

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 $x^3 + 1/x^3$, $x^4 + 1/x^4$.