Consider the following problem:
Each square of a $(2^n-1) \times (2^n-1)$ board contains either $1$ or $-1$. Such an arrangement is called successful if each number is the product of its neighbors. Find the number of successful arrangements.
Replace all $-1$s for $1s$ and all $1$s for $0$s and let $a_{i,j}$ be the new value the tile $(i,j)$ has. We can rewrite the initial condition as $$a_{i,j}+a_{i-1,j} + a_{i,j-1}+a_{i,j+1} + a_{i+1,j} \equiv 0 \mod{2}$$ for entries $a_{i,j}$ not at the edge nor at a corner, for those we can write similar equations with three or two summands respectively.
We now have $(2^n-1)^2$ equations and $(2^n-1)^2$ unknowns so let $A$ be the coefficient matrix with entries in $\mathbb{F}_2$. I am trying to show that the only board that satisfies the initial condition is the trivial one (the one in which all entries are $1$) and that is equivalent to proving that $\det(A) = 1$ in $\mathbb{F}_2$; however, I am not sure how to prove this.
Question: How do I prove that such matrix $A$ has determinant $1$ in $\mathbb{F}_2$?
Remark: Note that $A$ is not the initial board nor the one with updated values, it is instead the $(2^n-1)^2 \times (2^n-1)^2$ coefficient matrix for the system of linear equations in $\mathbb{F}_2$ I got by using the condition given. I am trying to prove that there is only a unique board that satisfies the condition, so this is why it is equivalent to proving that its determinant is odd in $\mathbb{R}$ or $1$ in $\mathbb{F}_2$.
Edit: This problem appears in a algebraic combinatorics handout by Yufei Zhao so the solution probably involves some technique mentioned in it.
Progress made: After pondering for a while and reading the handout over and over I came up with something.
Firstly, note that all algebra is being done in $\mathbb{F}_2$. We want to prove that the system of equations has a unique solution, this happens if and only if its matrix is invertible and this occurs if and only if the determinant is $1$. Now, note that if $a_{i,j}$ are the entries of the matrix $A$ and $S_{2^n-1}$ the permutations of $\{ 1, \ldots, 2^{n}-1 \}$ $$\det(A) = \sum_{\sigma \in S_{2^n-1}}(-1)^{\text{sgn } \sigma} a_{1, \sigma(1)}\ldots a_{2^n-1 , \sigma(2^n-1)} \equiv \sum_{\sigma \in S_{2^n-1}} a_{1, \sigma(1)}\ldots a_{2^n-1 , \sigma(2^n-1)}$$So we can conclude that $\det(A)$ is $1$ if and only if the permutations of the board that map each tile to a neighbour or to themselves is odd. Furthermore, after playing around with the $3 \times 3$ case I conjectured that there is only a single permutation that satisfies such condition, the identity permutation. Were this to be true the problem would be solved so I would appreciate any further help to move towards proving this last conjecture.
Let $A$ be the adjacency matrix for the $(2^n-1)\times (2^n-1)$ grid graph, where each vertex is adjacent to itself and its orthogonal neighbors. It suffices to prove that $\det A\equiv 1\pmod 2$. To do this, we use the representation $$ \det A\equiv \sum_{\pi \in S_{N}} a_{1,\pi(1)}a_{2,\pi(2)}\cdots a_{N,\pi(N)}\pmod 2, $$ where $N=(2^n-1)^2$.
Note that the quantity $a_{1,\pi(1)}\cdots a_{N,\pi(N)}$ is zero, unless the permutation $\pi$ satisfies the property that $v$ is always adjacent to $\pi(v)$ for each vertex $v$ in the grid. I will call such a permutation valid; a valid permutation is equivalent to a partition of the grid into oriented cycles. Each cycle can be a singleton, a transposition between orthogonally adjacent squares, or a cycle with length four or more.
To simplify, let us find some large classes of valid permutations with even cardinality. These can be ignored when determining the parity of the number of valid permutations.
From now on, instead of permutations, we will think of tilings with squares and dominos.
Next, we pair up each tiling with the result of reflecting that tiling though the vertical axis of symmetry of the grid. This shows that there are an even number of tilings without vertical symmetry.
Of the tilings with vertical symmetry, we apply the same method with symmetry across the horizontal axis. Now, the only remaining tilings have both vertical symmetry and horizontal symmetry.
A tiling with vertical and horizontal symmetry is uniquely specified by the tilings in three subregions. Subregion 1 is the upper-left $(2^{n-1}-1)\times (2^{n-1}-1)$ sub-matrix. By induction, subregion 1 can be tiled in an odd number of ways. The horizontal and vertical symmetry then forces the tilings in the $(2^{n-1}-1)\times (2^{n-1}-1)$ subsquares in the upper-right, lower-left, and lower-right corners.
If we ignore these four $(2^{n-1}-1)\times (2^{n-1}-1)$ sub-matrices, then what remains is a "cross", consisting of four "arms" of length $2^{n-1}-1$, surrounding the central square. Subregion $2$ is the leftward arm of the cross. This is a $(2^{n-1}-1)\times 1$ rectangle, and we must show that there are an odd number of ways to tile it with squares and dominoes (see next paragraph for the proof). Finally, subregion 3 is the upper arm of the cross, which is congruent to subregion 2. The tilings in subregions 2 and 3 determine the tilings in the rightward and lower arms of the cross, by symmetry.
To prove that a $(2^{n-1}-1)\times 1$ rectangle can be tiled in an odd number of ways by squares and dominoes, we treat this as a separate problem, and proceed by induction. There are an even number of tilings which are NOT symmetric when rotated $180^\circ$ about the central square, so we can ignore these. The remaining symmetric tilings are uniquely determined by the tiling of the left half of the rectangle, which is a $(2^{n-2}-1)\times 1$ rectangle. By induction, there are an odd number of ways to tile this, completing the proof of the sub-problem.
Putting this altogether, we see that the number of valid permutations has the same parity as the number of symmetric tilings, which are determined by independent tilings of three subregions, each of which can be done in an odd number of ways. Therefore, the number of valid permutations is odd, so the determinant in nonzero. QED!