Recurrence relation equations Suppose for all integers $n \ge 1$, 
$$x_{n+1}= 3x_n + 4y_n$$ and $$y_{n+1} = 2x_n + 3y_n$$ 
where $x_1 = 3, y_1 = 2$.
If $A(x_n)^2 + Bx_ny_n + C(y_n)^2 = 0$  for all integers $n \ge 1$, where $A$, $B$ and $C$ are constants, do $A$, $B$ and $C$ all have to be $0$?
 A: Let $X_n=\begin{pmatrix}x_n\\ y_n\end{pmatrix}$ and $M=\begin{pmatrix}3&4\\ 2&3\end{pmatrix}$, then $X_1=\begin{pmatrix}3\\ 2\end{pmatrix}$ and $X_{n+1}=MX_n$ for every $n$ hence $X_{n+1}=M^nX_1$. Likewise, $Ax_n^2+Bx_ny_n+Cy_n^2=X_n^TKX_n$ where $K=\begin{pmatrix}A&\frac12B\\ \frac12B&C\end{pmatrix}$.
If $Ax_n^2+Bx_ny_n+Cy_n^2=0$ for every $n\geqslant1$, then $Q_n=0$ for every $n\geqslant0$, where $$
Q_n=X_1^T(M^n)^TKM^nX_1.
$$
The matrix $M$ has eigenvalues $\lambda\gt\mu\gt0$ and $X_1$ is not an eigenvector of $M$ hence $X_1=U+V$ where $U$ and $V$ are eigenvectors of $M$ for the eigenvalues $\lambda$ and $\mu$ respectively. Thus, $M^nX_1=\lambda^nU+\mu^nV$ and 
$$
Q_n=\lambda^{2n}U^TKU+2\lambda^n\mu^n U^TKV+\mu^{2n}V^TKV.
$$
Since $\lambda^{2n}\gg\lambda^n\mu^n\gg\mu^{2n}$ when $n\to\infty$, this implies that 
$$
U^TKU=U^TKV=V^TKV=0.
$$
Thus, for every $(a,b)$, $(aU+bV)^TK(aU+bV)=0$, that is, for every vector $W$, $W^TKW=0$. In particular, $K$ is the matrix of the quadratic form which is identically zero, hence $K$ is the zero matrix, that is, $A=B=C=0$.

Key-arguments: The matrix $M$ is diagonalizable, its eigenvalues are nonzero and with distinct modulus, and $X_1$ is not an eigenvector of $M$.

A: One native way is to eliminate $y$ and  find $x_n$  and vice versa as follows:
From the first equation, we have $\displaystyle y_n=\frac{x_{n+1}-3x_n}4\implies y_{n+1}=\frac{x_{n+2}-3x_{n+1}}4$
Put these values of $y_n,y_{n+1}$ in the second equation to form an equation exclusively in $x_{n+2},x_{n+1},x_n$ and then use this method
