Let \(k\) be a positive integer. Determine all real numbers \(c\) such that every real sequence \((x_n)\) satisfying \[ \frac12(x_{n+1}+x_{n-1})=cx_n \] has period \(k\), i.e. \[ x_{n+k}=x_n \] for every \(n\).
Solution
Set \[ A_c\coloneqq \begin{pmatrix} 2c&-1\\ 1&0 \end{pmatrix}. \]
1. If\[ v_n\coloneqq \begin{pmatrix} x_n\\x_{n-1} \end{pmatrix}, \]then the recurrence is equivalent to\[ v_{n+1}=A_cv_n. \]
The recurrence may be rewritten as \[ x_{n+1}=2cx_n-x_{n-1}. \] Thus \[ \begin{pmatrix} x_{n+1}\\x_n \end{pmatrix} = \begin{pmatrix} 2c&-1\\1&0 \end{pmatrix} \begin{pmatrix} x_n\\x_{n-1} \end{pmatrix}. \]
2. Every solution of the recurrence satisfies\[ x_{n+k}=x_n \]for every \(n\) if and only if\[ A_c^k=I. \]
If \(A_c^k=I\), then step 1 gives \[ v_{n+k}=A_c^kv_n=v_n, \] so in particular \(x_{n+k}=x_n\).
Conversely, the initial pair \((x_1,x_0)\) may be chosen arbitrarily in \(\RR^2\). If every resulting solution satisfies the period condition, then \[ A_c^k \begin{pmatrix} x_1\\x_0 \end{pmatrix} = \begin{pmatrix} x_{k+1}\\x_k \end{pmatrix} = \begin{pmatrix} x_1\\x_0 \end{pmatrix} \] for every vector in \(\RR^2\). Hence \(A_c^k=I\).
3. The characteristic polynomial of \(A_c\) is\[ q_c(\lambda)=\lambda^2-2c\lambda+1. \]
Directly, \[ \det(\lambda I-A_c) = \det \begin{pmatrix} \lambda-2c&1\\ -1&\lambda \end{pmatrix} = \lambda^2-2c\lambda+1. \]
4. If \(A_c^k=I\), then\[ c=\cos\left(\frac{2\pi m}{k}\right) \]for some integer \(m\) with\[ 1\le m<\frac{k}{2}. \]
If \(A_c^k=I\), every eigenvalue \(\lambda\) of \(A_c\) satisfies \(\lambda^k=1\). Since \[ \det A_c=1, \] the two eigenvalues are \(\lambda\) and \(\lambda^{-1}\). Thus for some integer \(m\), \[ \lambda=e^{2\pi i m/k}, \] and step 3 gives \[ 2c = \lambda+\lambda^{-1} = 2\cos\left(\frac{2\pi m}{k}\right). \]
The cases \(\lambda=1\) and \(\lambda=-1\) cannot occur. Indeed, \(A_c^k=I\) implies that the minimal polynomial of \(A_c\) divides \(t^k-1\), which has no repeated roots over \(\CC\), so \(A_c\) is diagonalizable. If \(\lambda=1\) were the only eigenvalue, diagonalizability would force \(A_c=I\), impossible because the lower-left entry of \(A_c\) is \(1\). Similarly, if \(\lambda=-1\) were the only eigenvalue, diagonalizability would force \(A_c=-I\), again impossible.
Replacing \(m\) by \(k-m\) does not change the cosine, so every admissible value has a representative with \[ 1\le m<k/2. \]
5. Conversely, if\[ c=\cos\left(\frac{2\pi m}{k}\right) \]for an integer \(m\) with \(1\le m<k/2\), then\[ A_c^k=I. \]
Put \[ \theta=\frac{2\pi m}{k}. \] Since \(0<\theta<\pi\), the two roots of the characteristic polynomial in step 3 are distinct: \[ e^{i\theta} \qquad\text{and}\qquad e^{-i\theta}. \] Hence \(A_c\) is diagonalizable over \(\CC\). Both eigenvalues have \(k\)th power \(1\), so after diagonalization the \(k\)th power of \(A_c\) is the identity. Therefore \(A_c^k=I\).
6. The required real numbers are exactly\[ \boxed{ c=\cos\left(\frac{2\pi m}{k}\right), \qquad m=1,\ldots,\left\lfloor\frac{k-1}{2}\right\rfloor}. \]For \(k=1\) or \(k=2\), this set is empty.
Step 4 gives necessity, and step 5 gives sufficiency. The integers satisfying \(1\le m<k/2\) are exactly \[ 1,\ldots,\left\lfloor\frac{k-1}{2}\right\rfloor. \]
7. Q.E.D.
Step 6 is the requested classification.