Determinants: periodic entries $0,1,2,3$ Consider an $n\times n$ matrix $M_n$ where the sequence 
$$\{1,2,3,\dots,n^2\} \mod 4=\{1,2,3,0,1,2,3,\dots\}$$ forms a clock-wise spiral, in that given order. For example,
$$M_4=\begin{bmatrix} 1&2&3&0\\ 0&1&2&1\\ 3&0&3&2 \\ 2&1&0&3 \end{bmatrix} \qquad \text{and} \qquad
M_5=\begin{bmatrix} 1&2&3&0&1\\ 0&1&2&3&2 \\ 3&0&1&0&3 \\
2&3&2&1&0 \\ 1&0&3&2&1
\end{bmatrix}.$$ 

Question. Is it true that 
  $$\det(M_{2n})=3(2n-1)4^{n-1} \qquad \text{and} \qquad
\det(M_{2n+1})=-(3n^2-1)4^n\,\,\,?$$

Added clarification. To understand the construction of the above matrices, take a look at the matrices from my other MO question. Then, reduce the entries modulo $4$ and follow through by computing the determinants.
 A: Yes, it is true. More generally, the entries $1,2,3,0$ can be replaced by arbitrary numbers $a,b,c,d$, in which case the determinant of $M_n$ can be computed in terms of the four numbers $u = d-b$, $v = a-c$, $U = d+b$ and $V = a+c$ as follows:


*

*If $n=4k$ for some positive integer $k$, then
$$
\det\left(  M_{n}  \right)  =\dfrac{1}{4}
v^{n-4}\left(  v^{4}-u^{2}v^{2}+\left(  U^{2}-V^{2}\right)  \left(  \left(
2k-1\right)  ^{2}v^{2}-\left(  2k\right)  ^{2}u^{2}\right)  \right)  .
$$

*If $n=4k+2$ for some positive integer $k$, then
$$
\det\left(  M_{n} \right)  =-\dfrac{1}{4}%
v^{n-4}\left(  v^{4}-u^{2}v^{2}+\left(  U^{2}-V^{2}\right)  \left(  \left(
2k+1\right)  ^{2}v^{2}-\left(  2k\right)  ^{2}u^{2}\right)  \right)  .
$$

*If $n=4k+1$ for some positive integer $k$, then
$$
\det\left(  M_{n}  \right)  =\dfrac{1}{2}
u^{n-3}\left(  u^{2}\left(  v+V\right)  -\left(  2k\right)  ^{2}v\left(
U^{2}-V^{2}\right)  \right)  .
$$

*If $n=4k+3$ for some positive integer $k$, then
$$
\det\left(  M_{n}  \right)  =\dfrac{1}{2}
vu^{n-3}\left(  u^{2}+vV-\left(  2k+1\right)  ^{2}\left(  U^{2}-V^{2}\right)
\right)  .
$$
As you can imagine, this is not very fun to prove. I have a writeup (The 4-periodic spiral determinant) in which I attempt at making the idea clear without going into all the details; in particular, annoying computations are relegated to SageMath and to the reader (and on some occasions to a combination of both). Even at that level of terseness, it is 24 pages long. I would normally hope that something nicer can be found, but with the complexity of the answer I am not too hopeful.
The proof starts out as suggested by @user44191 in one of the comments to the original post; thus the matrix is brought to a form where all entries are zero except for those in northwesternmost $4\times 4$-submatrix and on four sub-antidiagonals (namely, the $1$-st, the $3$-rd, the $5$-th and the $7$-th sub-antidiagonals) below the main antidiagonal. Then, I turn the matrix upside down, so that the sub-antidiagonals become the super-diagonals. I then perform Laplace expansion with respect to the last $4$ rows. All $4 \times 4$-minors from the last $4$ rows can be explicitly computed (only $\dbinom{7}{4}$ of them nonzero, and this can be further reduced by looking at the vanishing of the complementary minors), so it remains to compute the complementary $\left(n-4\right)\times\left(n-4\right)$-minors. For this, Jacobi's complementary minor theorem turns out to be of use, along with an explicit computation of the inverse of a certain power series.
