This is a generalization of Prove by induction that $73\mid 8^{(n+2)}+9^{(2n+1)}$ which asks how to prove that $73$ divides $8^{n+2}+9^{2n+1} $for all $n$.
Here is my generalization: Find conditions on positive integers $u, a, v, c$ and non-negative integers $b, d$, such that there exists an $m$ which divides $f(n) = u^{an+b}+v^{cn+d}$ for all $n$. Determine $m$ in terms of $u, a, b, v, c, d$.
Here is what I have done so far:
My solution is to write $f(n+1)$ in terms of $f(n)$ and see what happens.
$\begin{array}\\ f(n+1) &= u^{a(n+1)+b}+v^{c(n+1)+d}\\ &= u^{an+a+b}+v^{cn+c+d}\\ &= u^a u^{an+b}+v^c v^{cn+d}\\ &= (u^a-v^c+v^c) u^{an+b}+v^c v^{cn+d}\\ &= (u^a-v^c) u^{an+b}+v^c (u^{an+b}+v^{cn+d})\\ &= (u^a-v^c) u^{an+b}+v^c f(n)\\ \end{array} $
Therefore, if $(u^a-v^c)|f(n)$ for some $n$, it divides $f(m)$ for all $m \ge n$.
Since $f(0) = u^b+v^d$, if $(u^a-v^c)|(u^b+v^d)$, it divides $f(n)$ for all $n$.
For the original problem, $(u, a, b, v, c, d) =(8, 1, 2, 9, 2, 1) $, so $u^a-v^c =8^1-9^2 =-73$ and $u^b+v^d =8^2+9^1 =73$, so $73|f(n)$ for all $n$.
To replicate this more generally, we want (assuming $v^c > u^a$) $v^c-u^a = u^b+v^d$ or $v^c-v^d = u^a+u^b$. For this to hold, we must have $c > d$ and, if $A = max(a, b)$ and $B = min(a, b)$, $v^d(v^{c-d}-1) = u^B(u^{A-B}+1) $. If $(u, v) = 1$, this implies that $v^d = u^{A-B}+1$ and $u^B = v^{c-d}-1$.
Writing the first of these as $v^d u^B = u^A+u^B$ and substituting the second, we get $v^d (v^{c-d}-1) = u^A+v^{c-d}-1$ or $v^d (v^{c-d}-1) = u^A+v^{c-d}-1$ or $u^A = v^{c-d}(v^d-1)-v^d+1 =(v^{c-d}-1)(v^d-1)$
(Now I'm playing around, looking for solutions.)
If $c-d = d$ (or $c = 2d$), we get $u^A = (v^d-1)^2$, so a solution is always $A=2$ and $u = v^d-1$. Since $u^B = v^{c-d}-1 =v^d-1 = u$, $B = 1$.
If $c = d+1$, $u^A = v(v^d-1)-v^d+1 = (v-1)v^d-v+1 =(v-1)(v^d-1)$. If $d=1$, a solution to this is always $u = v-1$ and $A = 2$.
That's all I've got.
