Let $A(x) = \prod\limits_i (x-\lambda_i)$ and $B(x) = \prod\limits_j (x-\mu_j)$. Then, their composed product is defined as $$ (A*B)(x) = \prod\limits_{i,j} (x-\lambda_i \mu_j). $$ Generally, we can show that if $A(x)$ and $B(x)$ have integer coefficients, so does $(A*B)(x)$. The reason for this is that if $A(x)$ and $B(x)$ are characteristic polynomials of matrices $A$ and $B$, then $(A*B)(x)$ is the characteristic polynomial of $A \otimes B$, their Kronecker product.

Now, let's focus on the composed product of $A(x)$ with itself. We can define polynomials $A_d(x)$ and $A_s(x)$ as

$$\begin{align} A_s(x) &= \prod\limits_{i < j} (x-\lambda_i \lambda_j), \\ A_d(x) &= \prod\limits_i (x-\lambda_i^2), \end{align}$$ in which case $(A*A)(x) = A_d(x) A_s^2(x)$. It appears that $A_s(x)$ and $A_d(x)$ should also have integer coefficients if $A(x)$ is as such, but I don't see any simple way to show it, or compute them without finding the roots, like it was possible with $(A*B)(x)$. Any ideas on how to proceed?

Ideally, I want to not only show that $A_s(x)$ and $A_d(x)$ have integer coefficients, but also provide some algorithmic way to find them if $A(x)$ is known, without finding all complex roots of $A(x)$.

The reason I ask is, assume that $A(x)$ is the characteristic polynomial of a linear recurrent sequence $u_n$. Then, $A_s(x)$ is the characteristic polynomial of the linear recurrent sequence $h_n = u_n u_{n+a+b} - u_{n+a} u_{n+b}$, and more generally, $A_s(x) A_d(x)$ appears to be the characteristic polynomial of $h_n = u_n u_{n+a}$, so it would be nice to have a general method to find it directly.