Let $A^n-I=\begin{bmatrix} a_{n}&b_{n}\\ c_{n}&d_{n} \end{bmatrix} $, where $A=\begin{bmatrix} 3&2\\ 4&3 \end{bmatrix}$, $I=\begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}$ and let $f_{n}=\gcd(a_{n},b_{n},c_{n},d_{n})$.
Show that $$\lim_{n\to\infty}f_{n}=\infty$$
My idea: I think we first have to find $A^n$, and this has a nice solution? Thank you.
We have the equation $A=B^2$ where $$ B=\left(\begin{array}{cc}1&1\\2&1\end{array}\right). $$
Let $$ B^n=\left(\begin{array}{cc}x_n&y_n\\z_n&x_n\end{array}\right). $$ As $\det B^n=(-1)^n$, we know that $x_n^2=z_ny_n+(-1)^n.$ An easy induction also proves that $z_n=2y_n$ for all $n$.
Assume first that $n$ is odd. I claim that $2x_n$ then divides all the entries of $A^n-I$. This is because $$ A^n=(B^n)^2= \left(\begin{array}{cc}x_n^2+y_nz_n&2x_ny_n\\ 2x_nz_n&x_n^2+y_nz_n\end{array}\right) $$ Here $y_nz_n=x_n^2+1$, so we get that $$ A^n-I=\left(\begin{array}{cc}2x_n^2&2x_ny_n\\ 2x_nz_n&2x_n^2\end{array}\right) $$ proving the claim.
Assume then that $n$ is even. This time I claim that all the entries of $A^n-I$ are divisible by $2y_n$. For even $n$ we have $y_nz_n=x_n^2-1$, so the above calculation shows that $$ A^n-I=\left(\begin{array}{cc}2y_nz_n&2x_ny_n\\ 2x_nz_n&2y_nz_n\end{array}\right) $$ proving the claim.
It is easy to prove that both $x_n,y_n\to\infty$ as $n\to\infty$. Your claim follows from this.
It seems that $$b_n=a_{n-1}+b_{n-1}+c_{n-1}+d_{n-1},~~c_n=2b_n,~~a_n=d_n=\sqrt{1+b_nc_n}$$
And for $a_n$ see A001541, for $b_n$ see A001542. I hope besides to complete elegant answer of @Jyrki; this help you also.
