Find the determinant of $$A=\begin{bmatrix} I_{N \times N} & Q_{N \times N} \\ Q_{N \times N} & I_{N \times N} \end{bmatrix},$$ where $$Q=\begin{bmatrix} q_1 & q_2 & q_3 & \ldots & q_N\\ q_{N+1} & q_1 & q_2 &\ldots & q_N-1\\ q_{N+2} & q_{N+1} & q_1&\ldots & q_N-2\\ \vdots & \vdots &\vdots &\ddots\\ q_{2N-1} & q_{2N-2} & q_{N-3}&\ldots & q_1\\ \end{bmatrix}$$.
We can use the following properties:
$$ \det {\begin{pmatrix}X&Y\\Y&X\end{pmatrix}}=\det(X-Y)\det(X+Y)$$ $$\det {\begin{pmatrix}X&Y\\S&T\end{pmatrix}}=\det(T)\det \left(X-YT^{-1}S\right)$$ To show that
$$|A|=|I-Q||I+Q|=|I-Q^2|.$$ Now, I want to use the properties of the matrix $Q$ (look at repeated elements and the fact that the anti-diagonal transpose of $Q$ is equal to $Q$) to simplify the result and write $|A|$ in terms of $q_1,\ldots,q_{2N}$. Any idea how to continue from here?