The Chebyshev polynomials of the first kind satisfy $T_m \cdot T_n = \frac{1}{2}(T_{n+m} + T_{n-m})$ for $0 \leqslant m \leqslant n$, and $T_m \cdot U_{n-1} = \frac{1}{2}(U_{n+m-1} + U_{n-m-1})$ for $0 \leqslant m < n$. For $m$ and $n$ natural integers such that $m \leqslant n$, we propose to determine the quotient $Q_{n,m}$ and the remainder $R_{n,m}$ of the Euclidean division of $T_n$ by $T_m$. a) Suppose $m < n < 3m$. Show that $$Q_{n,m} = 2T_{n-m} \quad \text{and} \quad R_{n,m} = -T_{|n-2m|}$$ b) Determine $Q_{n,m}$ and $R_{n,m}$ when $n$ is of the form $(2p+1)m$ with $p \in \mathbb{N}^*$. c) Suppose that $m > 0$ and that $n$ is not the product of $m$ by an odd integer. Show that there exists a unique integer $p \geqslant 1$ such that $|n - 2pm| < m$ and that $$Q_{n,m} = 2\left(T_{n-m} - T_{n-3m} + \cdots + (-1)^{p-1} T_{n-(2p-1)m}\right) \quad \text{and} \quad R_{n,m} = (-1)^p T_{|n-2pm|}$$
The Chebyshev polynomials of the first kind satisfy $T_m \cdot T_n = \frac{1}{2}(T_{n+m} + T_{n-m})$ for $0 \leqslant m \leqslant n$, and $T_m \cdot U_{n-1} = \frac{1}{2}(U_{n+m-1} + U_{n-m-1})$ for $0 \leqslant m < n$.
For $m$ and $n$ natural integers such that $m \leqslant n$, we propose to determine the quotient $Q_{n,m}$ and the remainder $R_{n,m}$ of the Euclidean division of $T_n$ by $T_m$.
a) Suppose $m < n < 3m$. Show that
$$Q_{n,m} = 2T_{n-m} \quad \text{and} \quad R_{n,m} = -T_{|n-2m|}$$
b) Determine $Q_{n,m}$ and $R_{n,m}$ when $n$ is of the form $(2p+1)m$ with $p \in \mathbb{N}^*$.
c) Suppose that $m > 0$ and that $n$ is not the product of $m$ by an odd integer. Show that there exists a unique integer $p \geqslant 1$ such that $|n - 2pm| < m$ and that
$$Q_{n,m} = 2\left(T_{n-m} - T_{n-3m} + \cdots + (-1)^{p-1} T_{n-(2p-1)m}\right) \quad \text{and} \quad R_{n,m} = (-1)^p T_{|n-2pm|}$$