Prime divisors of the sequence terms $a_n=a\cdot 2017^n+b\cdot 2016^n$ I am dealing with the test of the OBM (Brasilian Math Olimpyad), University level, 2017, phase 2.
As I've said at another topic (question 1), I hope someone can help me to discuss this test.
The question 2 says:

Taking fixed positive integers $a$ and $b$, show that the set of the prime divisors of the sequence terms $a_n=a\cdot 2017^n+b\cdot 2016^n$ is infinite.

The only thing that is on my mind is Dirichlet's Theorem: Given any $k,k'\in\mathbb{Z}$ coprime, the arithmetic progression of reason k' and inicial term k has infinite primes.
However, I don't have ideas about how do it. Thanks very much. 
Edit September, 01
I was searching about recurrences and I found a little about Lucas sequences, it seems important:
Lucas sequence
 A: Suppose the set of primes dividing the sequence is finite instead, and number them $p_1,...,p_K.$ 
Associate to each term $a_n = a \cdot 2016^n + b \cdot 2017^n$ the index $i, 1 \le i \le K,$ such that biggest prime power that divides $a_n$ is  $p_i^{\beta_i(n)}$. 
By the pigeonhole principle, there is an index $i_0$ and $a_n,a_m,\,m>n,m-n \le K,$ such that $a_n,a_m$ are both associated to $i_0,$ for each $n.$ Therefore, letting $l = \min(\beta_{i_0}(n),\beta_{i_0}(m)),$ we get that 
$$ p_{i_0}^l | a\cdot 2016^n + b\cdot 2017^n, p_{i_0}^l | a \cdot 2016^m + b\cdot 2017^m.$$ 
The relations above imply, on the other hand, that 
$$ p_{i_0}^l | b\cdot(2016^{m-n} - 2017^{m-n}). $$ 
From the fact that the number of primes is bounded, we get that $p_{i_0}^{l \cdot K} \ge 2017^n.$ On the other hand, by the equation above, 
$$ p_{i_0}^l \le C \cdot 2017^K.$$ 
As $K$ is fixed, we see that 
$$ 2017^{\frac{n}{K} - K} \le C, \text{for infinitely many } n \in \mathbb{N}.$$ 
This leads to a contradiction by letting $n \to \infty.$ 
