Putnam 2015, Problem B-3

Let $S$ be the set of all $2 \times 2$ real matrices
$$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
whose entries $a,b,c,d$ (in that order) form an arithmetic progression. Find all matrices $M$ in $S$ for which there is some integer $k > 1$ such that $M^k$ is also in $S$.

Solution:
First solution: Any element of \(S\) can be written as \(M = \alpha A + \beta B\), where \(A = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\), \(B = \begin{pmatrix} -3 & -1 \\ 1 & 3 \end{pmatrix}\), and \(\alpha, \beta \in \mathbb{R}\). Note that \(A^2 = \begin{pmatrix} 4 & 4 \\ 4 & 4 \end{pmatrix}\) and \(B^2 = \begin{pmatrix} -24 & -8 \\ 8 & 24 \end{pmatrix}\) are both in \(S\), and so any matrix of the form \(\alpha A\) or \(\beta B\), \(\alpha, \beta \in \mathbb{R}\), satisfies the given condition.

We claim that these are also the only matrices in \(S\) satisfying the given condition. Indeed, suppose \(M = \alpha A + \beta B\) where \(\alpha, \beta \neq 0\). Let \(C = \begin{pmatrix} 1 & 1/\sqrt{2} \\ -1 & 1/\sqrt{2} \end{pmatrix}\) with inverse \(C^{-1} = \begin{pmatrix} 1/2 & -1/2 \\ 1/\sqrt{2} & 1/\sqrt{2} \end{pmatrix}\). If we define \(D = C^{-1}MC\), then \(D = 2\alpha \begin{pmatrix} 0 & \gamma \\ \gamma & 1 \end{pmatrix}\) where \(\gamma = -\beta\sqrt{2}/\alpha\). Now suppose that \(M^k\) is in \(S\) with \(k \geq 2\). Since \((1 - 1)A \begin{pmatrix} 1 \\ -1 \end{pmatrix} = (1 - 1)B \begin{pmatrix} 1 \\ -1 \end{pmatrix} = 0\), we have \((1 - 1)M^k \begin{pmatrix} 1 \\ -1 \end{pmatrix} = 0\), and so the upper left entry of \(C^{-1}M^kC = D^k\) is 0. On the other hand, from the expression for \(D\), an easy induction on \(k\) shows that \(D^k = (2\alpha)^k \begin{pmatrix} \gamma^2 p_{k-1} & \gamma p_k \\ \gamma p_k & p_{k+1} \end{pmatrix}\), where \(p_k\) is defined inductively by \(p_0 = 0\), \(p_1 = 1\), \(p_{k+2} = \gamma^2 p_k + p_{k+1}\). In particular, it follows from the inductive definition that \(p_k > 0\) when \(k \geq 1\), whence the upper left entry of \(D^k\) is nonzero when \(k \geq 2\), a contradiction.

Remark: A variant of this solution can be obtained by diagonalizing the matrix \(M\).