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Source: Valentio Iverson

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#1 h Apr 22, 2025, 3:27 AM

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Let $P(x)$ be a monic polynomial with complex coefficients such that there exist a polynomial $Q(x)$ with complex coefficients for which $P(x^2-2)=P(x)Q(x).$ Determine all complex numbers that could be the root of $P(x)$ .

Proposed by Valentio Iverson, Indonesia

This post has been edited 1 time. Last edited by GreenTea2593, Apr 22, 2025, 3:27 AM

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#2 h Apr 22, 2025, 4:46 AM

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I don't know, is it full or fake solution?

Let $x=z$ be a root of $P(x)$ , then $x=z^2-2$ also a root of $P(x)$ . Let $u_n$ be a sequence such that
$u_{n+1} = (u_n)^2-2, u_1 = z$ , then all of elements of {u_n} are roots of $P(x)$ , so must be exist $m\in\mathbb{N}$ such that $u_m = u_1$ .

Let $g(x) = f^n(x) - x, f(x) = x^2-2$ , then $g(z) = 0$ , now we find all the value of $z$ , let $x= 2\cos{y}, y\in\mathbb{C}$ so we get $g(2\cos{y}) = 2(\cos(2^n y) - \cos y)$ , all solution of $y$ such that $g(2\cos{y}) =0$ are $y=0, y=\frac{2\pi m}{2^n\pm 1}$

This post has been edited 2 times. Last edited by Layaliya, Apr 22, 2025, 4:47 AM
Reason: Latex error

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#3 h Apr 22, 2025, 5:05 AM • 1 Y

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Layaliya wrote:

so must be exist $m\in\mathbb{N}$ such that $u_m = u_1$ .

This is not necessarily true, the sequence $u_1,u_2,\dots$ could be eventually periodic.

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#4 h Apr 22, 2025, 5:39 AM

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Let $f(x) = x^2 - 2.$
Suppose $z$ is a root of $P.$ Then, $f(z)$ must also be a root.
Thus, $f^i(z)$ are roots of $P$ for all $i = 0, 1, 2, 3, \dots.$
Hence, there must exist $m\in\mathbb N$ such that $f^m(z) = z,$ or else there are infinite roots.
Substitute $z = \alpha + \frac1\alpha$ and we get $\alpha^{2^m} + \frac1{\alpha^{2^m}} = \alpha + \frac1\alpha.$
Rearrangement gives $\alpha(\alpha^{2^m - 1} - 1) = \alpha^{-2^m}(\alpha^{2^m - 1} - 1)$ or $(\alpha^{2^m - 1} - 1)(\alpha^{2^{m} + 1} - 1) = 0.$
Thus, $\alpha$ are $2^m - 1$ or $2^m + 1$ roots of unity.
Thus, $z = 2\sin y$ where $y = \frac{2 \pi n}{2^m \pm 1}.$

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#5 h Apr 22, 2025, 5:55 AM

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Ilikeminecraft wrote:

Hence, there must exist $m\in\mathbb N$ such that $f^m(z) = z,$ or else there are infinite roots

Not necessarily true, there is a possibility that $f^m(z)=f^n(z)$ for positive integers $m>n$ .

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#6 h Apr 27, 2025, 7:57 AM

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Simple yet deadly lmao, @Greentea2593 orz. This is just a sketch, many details are omitted.

The answer is $2 \cdot cos (r\pi)$ for all rational number $r$ . First we'll prove that this is necesarry, the sufficiency will follow later.

We see that for every complex root of $P(x)$ , say $z$ . It also follows that $z^2-2$ is a complex root. Since $z$ is a complex number, there exist another complex number $x$ such that $z=x+\frac{1}{x}$ . Now by induction, we have $x^{2^{n-1}}+\frac{1}{x^{2^{n-1}}}$ is also a root for every natural number $n$ .

We see that $P(x)$ only has a finite amount of roots, hence the sequence $a_n=x^{2^{n-1}}+\frac{1}{x^{2^{n-1}}}$ must be eventually periodic.
Let's say for some natural numbers $m,n$ we have
$x^{2^{n-1}}+\frac{1}{x^{2^{n-1}}}=x^{2^{m+n-1}}+\frac{1}{x^{2^{m+n-1}}}$ Solving this will imply that $x$ is either the $2^{m+n-1}-2^{n-1}$ or $2^{m+n-1}+2^{n-1}$ 'th root of unity. Substitute this back to get $z=2\cdot cos\left( \frac{2i}{2^{m+n-1} \pm 2^{n-1}} \cdot \pi \right)$ where $i=0,1,\cdots ,2^{m+n-1} \pm 2^{n-1}-1$ .

Now we'll prove the sufficiency part. It suffices to prove for $r \in [0,1]$ , first notice that $\frac{2i}{2^{m+n-1} \pm 2^{n-1}}$ can take any rational value in $[0,1]$ for arbitrary natural number $m,n$ . Let $p=2^{m+n-1} \pm 2^{n-1}$ , we construct the polynomial
$P(x)= \prod_{i=0}^{p-1} \left( x- 2\cdot cos\left( \frac{2i}{p} \cdot \pi \right) \right).$ By using double angle identity we get

$\begin{align*} P(x^2-2) &= \prod_{i=0}^{p-1} \left( x^2-2- 2\cdot cos\left( \frac{2i}{p} \cdot \pi \right) \right)\\ &=\prod_{i=0}^{p-1} \left( x- 2\cdot cos\left( \frac{i}{p} \cdot \pi \right) \right)\left( x+ 2\cdot cos\left( \frac{i}{p} \cdot \pi \right) \right)\\ &=\prod_{i=0}^{p-1} \left( x- 2\cdot cos\left( \frac{i}{p} \cdot \pi \right) \right)\left( x- 2\cdot cos\left( \frac{p+i}{p} \cdot \pi \right) \right)\\ &=\prod_{i=0}^{2p-1} \left( x- 2\cdot cos\left( \frac{i}{p} \cdot \pi \right) \right) \end{align*}$ Hence $P(x)$ divides $P(x^2-2)$ . And the roots can take any number in the form of $2\cdot cos(r\pi)$ where $r$ is a rational number.

This post has been edited 5 times. Last edited by amogususususus, Apr 28, 2025, 5:02 AM

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