Assume wlog $(a,c)=1=(b,d)$ and name the sequence $x_n$. If $p|ab^{n_0}+cd^{n_0}$, and $t=ord_p(b/d)$ (note that if $p|(a,d)$ then $v_p(x_n)$ is bounded anyway, in which case we consider another prime; similarly if $p|(b,c)$), let us consider $v_p(x_{n_0+kt})$ This is $$v_p(d^{kt}(ab^{n0}+cd^{n0})+ab^{n0}(b^{kt}-d^{kt}))\geq \min\{v_p(ab^{n_0}+cd^{n_0});v_p(b^{kt}-d^{kt})\}$$with equality if the two valuations are different. So if the valuation of $k$ is big enough, and consequently by LTE the valuation of $b^{kt}-d^{kt}$ as well, then the valuation of $v_p(x_{n_0+kt})$ is bounded. If there is a prime $q$ dividing some $x_{n_0'}=x_{n_0+kt_0}$, with $ord_{q}(b/d)=t'$, then considering the same argument for $x_{n_0'+ktt'}$, shows that this is bounded for infinitely many $k$, by taking $v_p(k)$ and $v_q(k)$ big enough. Repeating this process, for all primes that can appear in these sequences (there can be finitely many), we eventually get to a sequence of the form $x_{N_0+kT}$, where $T=lcm(ord_{p_1}(b/d),...,ord_{p_m}(b/d))$, and if $v_{p_1}(k),...,v_{p_m}(k)$ are big enough, then the $p_i$-adic valuations of the $x_{N_0+kT}$ are themselves constant and so bounded, and say they are at most $C_i$. So, since there can be no other prime dividing this sequence[except those we excluded at the beginning for dividing $(a,d)$ or $(b,c)$, which also have bounded valuation], we must have $x_{N_0+kT}\leq p_1^{C_1}\cdots p_m^{C_m}$ which is clearly a contradiction, unless the original sequence is bounded. This can only happen if $b=d=1$ or in general $b=d$ of we multiply once again by the gcd of $b$ and $d$.

Taking ${\rm gcd}(b,d)$ outside if necessary, it suffices to show $b=d=1$ under $(b,d)=1$ Let $\{p_1,p_2,\dots,p_N\}$ be the primes dividing $a\cdot b^n + c\cdot d^n$. Assume the contrary that $b\cdot d>1$. Note that if $p_i\mid d$ then $p_i\mid a$. Consequently, \[ v_{p_i}\bigl(a\cdot b^n + c\cdot d^n\bigr) \]is bounded. Same also holds for any $p\mid b$ (which hence divides $c$). Let $\mathcal{I}\subset \{1,2,\dots,N\}$ with the property that $i\in \mathcal{I}$ iff $p_i\mid b\cdot d$. Now, take $L$ sufficiently large so that $\textstyle\min_{1\le i\le N}p_i^L\gg a+c$. Taking $n=\textstyle k\prod_{1\le i\le N}\varphi\bigl(p_i^L\bigr)$ (where $\varphi(\cdot)$ is Euler's function), it follows that for some $i\in \mathcal{I}^c$ (w.r.t. $\{1,2,\dots,N\}$), $p_i^L\mid a\cdot b^n+c\cdot d^n$. Indeed, since $b\cdot d\ne 1$, $a\cdot b^n+c\cdot d^n$ grows unbounded, and for those primes $p_i$ with $i\in\mathcal{I}$, their exponent is bounded. Notice, however, that this is a clear contradiction: \[ a\cdot b^n + c\cdot d^n\equiv a+c\pmod{p_i^L}, \]by Euler's theorem; but $p_i^L\gg a+c$. This completes the proof.

Similar to USA TSTST 2014/6 (which asks for a boundedness of $v_p$)

What is wrong with this proof? I wrote roughly the same in MO and got 0 Wlog $(a,c)=(b,d)=1$ $ab^n+cd^n \equiv 0 (\mod p) \iff (b/d)^n\equiv (-c/a) (mod p)$ Thus we want to prove that if $k^n-l$ has finitely many primes $k=1$. If $z$ is a big enough fixed number (for $p$'s with $p|k,l$) then for any $p|k^n-l$ for some $n$, we choose $\phi (p^a)|m-1$ for big enough $a$ (for other $p$'s) which shows $$v_p(k^{zm}-k^z)=v_p((k^{zm}-l)-(k^z-l))=min({v_p(k^zm-l),v_p(k^z-l)})$$ Because $a$ is big enough $v_p(k^{zm}-k^z) >v_p(k^z-l)$. Thus $v_p(k^{zm}-l) \le v_p(k^z-l)$. However $k^{zm}-l>k^z-l$ for $k>1$. Contradiction.$\square$ Maybe I should have said that there exists integer $k,l$ such that $k \equiv (b/d), l \equiv (-c/a) \mod (p_1p_2\cdot ... \cdot p_u)$.

WLOG,assume $(a,c),(b,d)=1$. Let $Q$ be the product of set of prime factors of the sequence,which dont divide $b$ or $d$.Let $Q=q_1q_2..q_k$.Now,let $M=Q^{2(ab+cd)}$. Then ,choose $s \equiv 1 (mod \phi(M))$.Then,we will have that $ab^s+cd^s \equiv ab+cd (mod M)$,so for any such $q_i$, $v_{q_i}{(ab^s+cd^s)}=v_{q_i}{(ab+cd)}$.For any other prime $r$(suppose it divides $b$,then it divides $c$),we have that for sufficiently big $s$,$v_r(ab^s+cd^s) = v_r(c)$.Hence for all sufficently large $s$,$v_p(ab^s+cd^s)$ is bounded ,which gives that the term itself is bounded which is a contradiction.

SerdarBozdag wrote: $p$ does not divide $d,a$ because of the fact $(a,c)=(b,d)=1$. Thus $d^{-1},a^{-1}$ exist (or again I am missing something). Edit: I realised that $p|a,d$ can be a problem. Edit.2: I see that the first post has an idea to fix this. So this not a big mistake, or is it? I think there might also be another problem. While there is $k\equiv b/d\pmod{(p_1\cdots p_u)^c)}$ for a fixed integer $c$ there's no positive integer which satisfies the relationship for all $c$, because that would imply that $k=b/d$, which in general can't be true if $d\nmid b$; the same problem applies for $l\equiv -c/a$. Therefore I'm not so sure you can apply p-adic valuation, if the numbers $k$ and $l$ are not fixed. Maybe you can apply the same argument by leaving $k$ and $l$ as fractions (of course making sure of not having any problem with denominators divisible by $p$), although you should verify that everything still holds for rational numbers. Anyway, I'm wondering if Kobayashi's theorem also works for sequences of rational numbers (where the divisors, both of the numerators and denominators, are in finite number).

SerdarBozdag wrote: @below I fixed $m$ and $z$ big enough ($z$ is fixed thus prime powers are fixed thus $a$'s can be fixed in $\phi{p^a}$ for all $p$ by taking maximum $a+1$) so the $c$ you wrote is fixed (or at least has an upper-bound). I see. I guess it can work then, although I'm not totally sure.

Kinda disappointing. It appears this is taken from Brazil Undergrad Math Olympiad, 2017.

WLOG $\gcd(b,d)=1$ and let $S=\{p_1 , \dots p_k\}$ be the set of primes dividing $\{ab^n + cd^n\}_{n=1}^\infty$, assume at least one of $b$, $d$ is greater than $1$. Strategy: The idea is that for large $n$ letting $ab^n + cd^n = \prod p_i^{a_i}$, some $a_i$ is sufficiently large and we want $n \equiv 0 \pmod{p_i^{\varphi(t)}}$ and $t \le a_i$, this way we have \[0 \equiv ab^n + cd^n \equiv a\{0,1\} + c \{0,1\} \pmod{p_i^t}\]meaning $b^n \equiv d^n \equiv 0 \pmod{p_i^t}$ (otherwise size contradiction) which still contradicts our coprime assumption. We follow the strategy, let $n = \prod p_i^{\varphi(2^{a_i})} = \prod p_i^{2^{a_i-1}}$ for sufficiently large integers $a_1 , \dots , a_k$ then some prime $p$ has $\nu_p(ab^n + cd^n)$ sufficiently large and let $x$ be another large enough natural such that $2^{x-1} \le 2^{a_i - 1}$ and $2^x \le \nu_p(ab^n + cd^n)$. Then, \[0 \equiv ab^n + cd^n \equiv a\{0,1\} + c\{0,1\} \pmod{p^{2^x}}\]and we get a contradiction as per our strategy.

we just take the biggest p^vp(x) and as the primes are finite so we call them p_1, p_2 to p_t and the t+1 number of neighbors must have two same p and it goes wrong

Here is my solution: https://calimath.org/pdf/TurkeyMO2021-5.pdf And I uploaded the solution with motivation to: https://youtu.be/UEOTpDpGh2g