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Let $F \in \mathbb{C}(X)$ be a rational function with complex coefficients (and $\mathbb{C}(X)$ the set of such functions) i.e. $F=\frac P Q$ for $P,Q \in \mathbb{C}[X]$

I need to show that

$F(X)=F(\frac 1 X) \iff \exists G \in \mathbb{C}(X), G(X+ \frac 1 X)=F$

Showing $\Leftarrow$ is easy but I don't know how to show $\Rightarrow$.

Can someone help me please ?

Thanks in advance.

Bastien Tourand
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  • here is a hint : first prove by induction that the statement is true for $X^n+\frac{1}{X^n}$ for all integers n. then generalize it for any $F \in \mathbb{C}(X)$ – Laassila souhayl Apr 18 '21 at 17:04
  • I already know that there exists a sequence of polynomials $(P_n)$ such that $P_0=2$, $P_1=X$ and $P_{n+1}=X P_{n}-P_{n-1}$ satisfying $P_n(X+\frac 1 X)=X^n+\frac{1}{X^n}$ but I don't know how to generalize... – Bastien Tourand Apr 18 '21 at 17:20
  • I will give you more hints, using the sequence above, show that for every polynomial $Q\in\mathbb{C}[X]$, the statement is true for $Q(X)Q(\frac{1}{X})$ and $Q(X)+Q(\frac{1}{X})$. – Laassila souhayl Apr 19 '21 at 14:05
  • next write $F=\frac{1}{2}(F(X)+F(\frac{1}{X}))$ with $F=\frac{P}{Q}$. – Laassila souhayl Apr 19 '21 at 14:08

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