Let $(a_n)_{n \in \mathbb{N}}$ be the sequence of integers defined recursively by $a_0 = 0$, $a_1 = 1, a_{n+2} = 4a_{n+1}+a_{n}$ for $n \geq 0$. Find the common divisors of $a_{1986}$ and $a_{6891}$.
I think it is true that $\gcd(a_m,a_n) = a_{\gcd(m,n)}$, but I am not sure how to prove it. From this the answer follows since $\gcd(1986,6891) = 3$ and so $\gcd(a_{1986,6891}) = a_3 = 17$.
Your $\,a_i\,$ satisfy the same addition law as the Fibonacci numbers (with the same easy proof as as in this answer using matrix multiplication). Therefore the short proof I gave in this answer shows that this sequence too is a strong divisibility sequence, i.e. $\, (a_m,a_n) = a_{\large (m,n)}\,$ which immediately yields the sought result.
Remark $\ $ Responding to comments, below are further details. As in the first link we have
$$ \begin{bmatrix}a_2 &a_1\\ a_1 & a_0\end{bmatrix} = \begin{bmatrix}4 &1\\ 1 & 0\end{bmatrix},\quad \begin{bmatrix} a_{n+2} &\!\!\! a_{n+1}\\ a_{n+1} & \!\!\!a_n\end{bmatrix} = \begin{bmatrix} a_{n+1} &\!\! a_{n}\\ a_{n} & \!\!\!\!a_{n-1}\end{bmatrix} \begin{bmatrix}4 &1\\ 1 & 0\end{bmatrix}$$
Thus we infer by induction
$$ A_n := \begin{bmatrix} a_{n+1} &\!\! a_{n}\\ a_{n} &\!\!\! a_{n-1}\end{bmatrix} = \begin{bmatrix}4 &1\\ 1 & 0\end{bmatrix}^n\! =\, A_1^n $$
Therefore we deduce that $\,A_{m+n} = A_1^{m+n} = A_1^m A_1^n = A_m A_n,\ $ i.e.
$$ \begin{align} \begin{bmatrix} a_{m+n+1} &\!\! a_{m+n}\\ a_{m+n} &\!\!\!\! a_{m+n-1}\end{bmatrix} &= \begin{bmatrix} a_{m+1} &\!\! a_{m}\\ \!a_{m} & \!\!\!\!a_{m-1}\end{bmatrix} \begin{bmatrix} a_{n+1} &\!\! a_{n}\\ a_{n} & \!\!\!\!a_{n-1}\end{bmatrix}\\[.5em] &= \begin{bmatrix}a_{m+1}a_{n+1}+a_m a_n &\! a_{m+1}a_n+a_m a_{n-1}\\ a_m a_{n+1}+a_{m-1}a_n &\! a_m a_n + a_{m-1} a_{n-1} \end{bmatrix}\\ \end{align} $$
This yields the addition law $\ a_{m+n} =\, a_{m+1} a_n +a_m a_{n-1}.\ $ For example
$$ \begin{align} \begin{bmatrix}a_8 & a_7\\ a_7 & a_6\end{bmatrix} &= \begin{bmatrix}a_4 & a_3\\ a_3 & a_2\end{bmatrix} \begin{bmatrix}a_5 & a_4\\ a_4 & a_3\end{bmatrix}\\[.4em] &= \begin{bmatrix}72 & \!17\\ 17 &\! 4\end{bmatrix} \begin{bmatrix}305 &\!\!\!\!72\\ 72 &\!\!\!\!\! 17\end{bmatrix} = \begin{bmatrix}23184 &\!\!\! 5473\\ 5473 &\!\!\! 1292\end{bmatrix} \end{align} $$
For $\,f_n = \dfrac{x^n-y^n}{x-y},\,$ satisfying $\,f_{n+2} = \overbrace{(x\!+\!y)}^{f_2} f_{n+1}-xy\, f_n,\,$ a very similar proof shows that it has an analogous addition law: $\, f_{n+m}\ =\ f_m\ f_{n+1}\,\ -\ xy\, f_{m-1} f_n,\ $ i.e.
$$ \dfrac{x^{n+m}\!-y^{n+m}}{x-y}\,=\, \dfrac{x^{m}\!-y^{m}}{x-y}\,\dfrac{x^{n+1}\!-y^{n+1}}{x-y} - xy\,\dfrac{x^{m-1}-y^{m-1}}{x-y}\,\dfrac{x^{n}-y^{n}}{x-y}\qquad$$
To help dispel doubts in the comments, here is an Alpha verification of the prior equation.
Using the characteristic polynomial (you might prove the following by induction) we get that $x^{n+2}=4x^{n+1}+x^n$, implying that $$a_n=\left(2+\sqrt{5}\right)^{n}+\left(2-\sqrt{5}\right)^n$$ so we can denote this with $\alpha^n+\beta^n.$
We first let $k$ be a nonnegative odd integer, and note that $a_n>0$, and that $\frac{a_kn}{a_n}$ is rational. Then by expanding, we obtain that
$$\frac{a_{kn}}{a_n}=\alpha^{(k-1)n}-\alpha^{(k-2)n}\beta^n+...+\beta^{(k-1)n}$$
which means that $\frac{a_{kn}}{a_n}$ is of the form $a+b\sqrt{5}$. Recalling that it is indeed rational, we get that it must be an integer, so $a_n \vert a_{kn}$.
In the case that $k$ is even,
$$\frac{a_{kn}}{a_n}=\frac{\alpha^{kn}+\beta^{kn}}{\alpha^n+\beta^n}=\sum_{i=0}^{k-1}(-1)^i\cdot\alpha^{k-1-i}\cdot\beta^k+2\beta^k$$
which is again of the form $a+b\sqrt{5}$ so we get a similar argument here.
So, we get that if $m \vert n$, then $a_m \vert a_n$.
