Solve equation for $x$ given by a determinant I am looking for some interesting solution for this problem. Given $a\in \mathbb{C}$, solve for $x$ the following equation
$$
\begin{vmatrix}
1 & a & a+x & a+x^2\\
a & 1 & a+x^2 & a+x \\
a + x & a+ x^2 & 1 & a \\
a +x^2 & a+x & a & 1 \\
\end{vmatrix} =0\qquad (D_a)
$$
Set $Z_a = \lbrace x\in \mathbb{C} : x \text{ satisfies }(D_a)\rbrace$ to be  the set of solutions of problem $D_a$.
Anyway, I am searching some trick or idea to solve the problem (i.e., find the set $Z_a$ for every value $a$). Obviously, I am not interested in computing explicitly the determinant, instead I wonder if there is a  simple and elegant way to solve the problem (maybe using properties of determinant).
Note that the matrix is symmetric. This could be helpful (there are some beautiful results concerning symmetrics matrix).
Thanks in advance. Any idea is welcome!
 A: This is a block matrix of the form $\pmatrix{A&B\\B&A}$. Thus, we have
$$
\det \pmatrix{1 & a & a+x & a+x^2\\
a & 1 & a+x^2 & a+x \\
a + x & a+ x^2 & 1 & a \\
a +x^2 & a+x & a & 1} = \\
\det \left[
\pmatrix{1&a\\a&1}^2 - \pmatrix{a+x&a+x^2\\a+x^2 & a+x}^2
\right] = \\
\det\left[
\pmatrix{1&a\\a&1} - \pmatrix{a+x&a+x^2\\a+x^2 & a+x}
\right]
\det \left[
\pmatrix{1&a\\a&1} + \pmatrix{a+x&a+x^2\\a+x^2 & a+x}
\right] = \\
\det
\pmatrix{1-a-x&-x^2\\-x^2&1-a-x}\cdot 
\det 
\pmatrix{1+a+x&2a+x^2\\2a+x^2&1+a+x}.
$$
We see that a matrix of the form $\pmatrix{a&b\\b&a}$ is singular iff $a = b$ or $a = -b$. Thus, we obtain the desired values of $x$ by solving the following equations:

*

*$1 - a - x = -x^2$

*$1 - a - x = x^2$

*$1 + a + x = 2a + x^2$

*$1 + a + x =  -2a - x$.


Another approach: in terms of Kronecker products, we could express your matrix as
$$
M = I + a I \otimes J + x J \otimes I + (a + x^2)J \otimes J,
$$
where
$$
I = \pmatrix{1&0\\0&1}, \quad J = \pmatrix{0&1\\1&0}.
$$
Thus, the eigenvalues of $M$ will be of the form $1 + a\lambda_1 + x\lambda_2 + (a + x^2) \lambda_1\lambda_2$, where each $\lambda_i$ is one of the eigenvalues of $J$ (which are $\pm 1$). $M$ has determinant $0$ iff one of these eigenvalues is $0$.

Another approach: we can diagoanlize the matrix by computing $U^TMU$, where
$$
U = \frac 12 \pmatrix{1&1\\1&-1} \otimes \pmatrix{1&1\\1&-1} = 
\frac 12 \pmatrix{1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1}.
$$
A: Looking at the matrix and the symmetries, we can expect four quadratic terms.
Done by hand (what I learnt so long time ago)
$$D(a)=\left(x^2-x-a+1\right) \left(x^2-x+a-1\right) \left(x^2+x+a-1\right) \left(
   x^2+x+3a+1\right)$$
