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Assume
$$a+\frac{1}{a}$$ is a rational number.
Prove that $a^n+\frac{1}{a^n}$ is a rational number.

I can easily prove it using induction. I would like to know if is it possible to prove it without induction.

We can see that:

$$a^2+\frac{1}{a^2}=\left(a+\frac{1}{a}\right)^2-2 \in \Bbb Q$$

$$a^3+\frac{1}{a^3}=\left(a+\frac{1}{a}\right)^3-3a^2\frac{1}{a}-3a\frac{1}{a^2}=\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)\in \Bbb Q$$

Maybe we can prove that: $$a^k+\frac{1}{a^k}=P\left(a+\frac{1}{a}\right)$$ where $P$ is a polynomial?

Edit It is not a duplicate of this question since there, the question asks if $a^n+\frac{1}{a^n}\in \Bbb Z$ when $a+\frac{1}{a}\in \Bbb Z$. I ask if $a^n+\frac{1}{a^n}\in \Bbb Q$ when $a+\frac{1}{a}\in \Bbb Q$

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    $\begingroup$ Your last statement can also be proved by induction, is that allowed? $\endgroup$
    – Zima
    Commented Jun 6 at 8:44
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    $\begingroup$ @Zima We don't know that. For rational $a$ the problem is easy. $\endgroup$
    – Piotr Wasilewicz
    Commented Jun 6 at 8:48
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    $\begingroup$ "Without induction" is a pretty hairy requirement, because induction creeps up quite often in these kinds of statements, and some times it's very cleverly hidden away. $\endgroup$
    – Arthur
    Commented Jun 6 at 8:48
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    $\begingroup$ You can use math.stackexchange.com/a/1584479/42969 pllus the fact that $\cos(n \theta)$ can be expressed as a polynomial in $\cos(\theta)$, see en.wikipedia.org/wiki/Chebyshev_polynomials $\endgroup$
    – Martin R
    Commented Jun 6 at 8:49
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    $\begingroup$ Proving a proposition like this for general $n$ is theoretically impossible with Peano axioms, because the induction axiom will be required. If you do it without induction, the induction is hidden in one of the results you use. $\endgroup$
    – Mark Bennet
    Commented Jun 6 at 9:24

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