Find the common divisors of $a_{1986}$ and $a_{6891}$

Solution 1:

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

$$ \begin{align} \begin{bmatrix}a_8 & a_7\\ a_7 & a_6\end{bmatrix} &= \begin{bmatrix}a_4 & a_3\\ a_3 & a_2\end{bmatrix} \begin{bmatrix}a_5 & a_4\\ a_4 & a_3\end{bmatrix}\\[.4em] &= \begin{bmatrix}72 & \!17\\ 17 &\! 4\end{bmatrix} \begin{bmatrix}305 &\!\!\!\!72\\ 72 &\!\!\!\!\! 17\end{bmatrix} = \begin{bmatrix}23184 &\!\!\! 5473\\ 5473 &\!\!\! 1292\end{bmatrix} \end{align} $$

Regarding $\,f_n = \dfrac{x^n-y^n}{x-y},\,$ which satisfies $\,f_{n+2} = (x\!+\!y) f_{n+1}-xy\, f_n,\,$ a similar proof as above shows that it satisfies the addition law $\, f_{m+n} = f_{m+1} f_n - xy\, f_m f_{n-1},\ $ i.e.

$$ \dfrac{x^{m+n}\!-y^{m+n}}{x-y}\,=\, \dfrac{x^{m+1}\!-y^{m+1}}{x-y}\,\dfrac{x^{n}\!-y^{n}}{x-y} - xy\, \dfrac{x^{m}-y^{m}}{x-y}\,\dfrac{x^{n-1}-y^{n-1}}{x-y}$$

To help dispel doubts in the comments, here is an Alpha verification of the prior equation.