Show this recursively defined integer sequence $a_{n+1} = (2b_1 - a_1)a_n + (a_1b_1 - b_1^2 - 2a_1^2)a_{n-1}$ is never $0$. Let $a_1, b_1 \in \mathbb{Z}$ be arbitary integers with $a_1 \neq 0$, and define sequences $\{a_n \}_{n \geq 1}$ and $\{b_n \}_{n \geq 1}$ to be solutions to the following linear system of recursions:
\begin{cases} 
      a_{n+1} = (b_1 - a_1)a_n + (a_1)b_n \\
      b_{n+1} = (-2a_1)a_n + (b_1)b_n
\end{cases}
I have done the following manipulations which give a recursive formula for $a_n$.
First, observe the first equation gives us $b_{n} = \dfrac{a_{n+1} - (b_1 - a_1)a_n}{a_1}$.
Substituting this into the second equation, we see $$b_{n+1} = (-2a_1)a_n + b_1 \Big(\dfrac{a_{n+1} - (b_1 - a_1)a_n}{a_1}\Big).$$
Reindexing $n + 1 \rightarrow n$, we achieve $$b_n = (-2a_1)a_{n-1} + b_1 \Big(\dfrac{a_{n} - (b_1 - a_1)a_{n-1}}{a_1}\Big).$$
Finally, substituting this back into the first equation, we get:
$$ a_{n+1} = (b_1 - a_1)a_n + a_1 \Big[(-2a_1)a_{n-1} + b_1 \Big(\dfrac{a_{n} - (b_1 - a_1)a_{n-1}}{a_1}\Big) \Big].$$
Simplifying, we find $$a_{n+1} = (2b_1 - a_1)a_n + (a_1b_1 - b_1^2 -  2a_1^2)a_{n-1}.$$
I suspect, and would like to prove, that $a_n \neq 0$ for all $n \geq 1$ and for all choices $a_1, b_1 \in \mathbb{Z}$ where $a_1 \neq 0$.
As corrected in the comments, $\lim\limits_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$ does not appear to be $-3$, so my claim of well-behavedness asymptotically isn't even true. Any suggestions/hints for forward movement are appreciated.
 A: Set $a_0=0$ and $b_0=1$. Then the original recurrence works for $n\ge0$. Write it in matrix form:
$$
\begin{pmatrix}
      a_{n+1} \\ b_{n+1} 
\end{pmatrix}
=
\begin{pmatrix}
      b_1 - a_1 & a_1 \\ -2a_1 & b_1 
\end{pmatrix}
\begin{pmatrix}
      a_{n} \\ b_{n} 
\end{pmatrix}
=
A
\begin{pmatrix}
      a_{n} \\ b_{n} 
\end{pmatrix}
$$
Then
$$
v_{n}=
\begin{pmatrix}
      a_{n} \\ b_{n} 
\end{pmatrix}
=
A^n
\begin{pmatrix}
      a_{0} \\ b_{0} 
\end{pmatrix}
=
A^n
\begin{pmatrix}
      0 \\ 1 
\end{pmatrix}
= A^n v_0
$$
Therefore
$$
A^n
=
\begin{pmatrix}
      * & a_n \\ * & b_n 
\end{pmatrix}
$$
If $a_n=0$, then $v_0$ is an eigenvector of $A^n$ with eigenvalue $b_n$. The eigenvectors of $A^n$ are the same as the eigenvectors of $A$, and these have nonzero first coordinates, unlike $v_0$.
The calculations were made with WA.
A: Write the sequence in the matrix form, i.e. let $A_n = <a_n,b_n>$ and let $A$ be matrix with entries $A[1,1] = b_1-a_1$, $A[1,2] = a_1$, $[2,1] = -2a_1$, and $A[2,2] = b_1$. Then the equations defining the sequence can be written as $A_{n+1} = A*A_n$.
By induction one gets that $A_{n+1}= A^n*A_1$. Now, if $A$ can be diagonalized (i.e. it is the case if it has two distinct eigenvalues), say $A= C*D*C^{-1}$ then we get $A_{n+1} = C*D^n*C^{-1}A_1$.
From this formula one can see that if $\lambda_1$ and $\lambda_2$ are eigenvalues of $A$ then $A_{n+1}$ is a combination of their powers, i.e. $$a_{n+1} = M*(\lambda_1)^{n} + N*(\lambda_2)^{n}$$ where $M,N$ are some constants. Similar formula holds for $b_n$ (with different constants). So the problem reduces to find eigenvalues of $A$ that is roots of $(b_1-a_1 - t)(b_1-t) + 2a_1b_1 = 0$ i.e. $$t*2 - (2b_1+ a_1)t +2a_1b_1 = 0$$ Once finds $\Delta$ for this equation to be
$\Delta = (2b_1+ a_1)^2-8a_1b_1 = (2b_1 -a_1)^2$.
Clearly, as long as $\Delta$ is non zero we get two distinct eigenvalues.
As remarked in one of the comments, if (for example) $a_1=2, b_1=1$ we may get zero.
