# Divisibility of Chebyshev Polynomials

I was trying to solve a problem involving an Insect crawling on the Cartesian/Coordinate Plane.

We have an insect on the origin of the coordinate plane, who remembers a particular angle $$\theta.$$ We can command the insect to move in terms of a given string of commands with $$\text{R's}$$ and $$\text{M's}.$$

The command $$R$$ is for "Rotate an angle $$\theta$$ anticlockwise", the command $$M$$ for "Move $$1$$ Step forward in whatever angle you are facing"

For example, if the insect remembers the angle $$\theta=90^{\circ},$$ and performs the string MRMRMRM, then it will trace a square path, and will finally get back to the origin.

I need to prove that if $$\theta$$ is an angle such that all of the numbers; $$\cos{\theta}$$ $$\cos(2\theta)$$ $$\cos(3\theta)$$ $$\vdots$$

are irrational, then there is no such string of $$\text{M's}$$ and $$\text{R's}.$$

But this is not really question I want help on, I have an argument that can reduce this to proving an easier fact. We let $$T_k$$ denotes the $$k^\text{th}$$ Chebyshev Polynomial $$\big($$so that $$T_k(\cos{\theta})=\cos{(k\theta)}\big),$$ and let $$n$$ be a natural number.

Lemma 1: We need to prove that, for all nonnegative integers $$a_0,a_1,\dots,a_n,$$ there exists some polynomial $$p(x),$$ and integers $$a, b,$$ and a Chebyshev Polynomial $$T_m$$ such that; $$p(x)\big(a_n T_n(x)+a_{n-1}T_{n-1}(x)+\cdots+a_0T_0(x)\big)=aT_{m}(x)+b.$$

This is what I need help on. I have tried proving this by Strong Induction, but it didn't help much :(

Can you help me prove Lemma 1, since I can manage the rest. If you think it is not true, you can tell me about it. In that case you can try to answer the original question.

Thank You and Regards,

MathEagle

For example, $$P(z) = 1 + z + z^2 + z^4 + z^6 + z^7 + z^8$$ is irreducible over the rationals, and has four roots on the unit circle and four not. Because of the roots off the unit circle, it does not divide any cyclotomic polynomial, and its roots can't be roots of unity. If $$r$$ is one of the roots on the unit circle, $$\theta = \log(r)/i$$ has the property that none of $$\cos(k\theta)$$ are rational. But with this $$\theta$$ the insect would come back to the origin for the string MRMRMRRMRRMRMRM.
EDIT: $$e^{i\theta} = r$$ is an algebraic integer, and so is its complex conjugate $$\overline{r}$$. Therefore $$2 \cos(k \theta) = r^k + \overline{r}^k$$ is also an algebraic integer. But the only algebraic integers that are rational numbers are (ordinary) integers. So the only possible rational values for $$\cos(k\theta)$$ would be $$0$$, $$\pm 1/2$$ or $$\pm 1$$. But that would make $$\exp(ik\theta)$$, and thus also $$\exp(i\theta)$$, a root of unity, which can't happen.