TITLE: Coprime sequences from a linear recurrence.
QUESTION [6 upvotes]: Given the linear recurrence $a_n=5a_{n-1}-2a_{n-2}$ and the initial conditions $x_0=0,x_1=1$ and $y_0=1, y_1=3$ we get the two sequences
$$(x_n)=0,1,5,23,105,\ldots\\
  (y_n)=1,3,13,59,269,\ldots$$
which seem to be coprime is the sense that $gcd(x_n,y_n)=1$ for every $n$ (I checked it numerically up to $n=100$).
To prove the conjecture I tried (without success) the induction principle and Bezout's lemma to generate numbers $r$ and $s$ with $rx_{n+1}+sy_{n+1}=1$ from the existence of numbers $p$ and $q$ with $px_n+qy_n=1$.
Second, I solved the recurrence with the result
$$x_n=\frac{\alpha^n-\beta^n}{\alpha-\beta},\quad y_n=\frac{1}{2}(x_n+\alpha^n+\beta^n)$$
where $\alpha$ and $\beta$ are the roots of the characteristic equation $x^2=5x-2$. Again, I don't see how to derive a proof. I'm grateful for any help.

REPLY [2 votes]: Following @Bill, write the recurrence in the form $x_{n+1}=\frac{x_{n+2}+2x_n}{5}$ ( and similarly $y_{n+1}$) to obtain
$$x_{n+1}y_n-y_{n+1}x_n=\frac{x_{n+2}y_n-x_ny_{n+2}}{5}.$$
So any prime $p\neq 5$ that divides both $x_{n+2}$ and $y_{n+2}$ divides $x_{n+1}y_n-y_{n+1}x_n$.
On the other hand, the solution of the recurrence yields
$$x_{n+1}y_n-y_{n+1}x_n=\alpha^n\cdot\beta^n=2^n.$$
So $p$ must be $2$, a contradiction, as $y_n$ is always odd. But also $p=5$ is impossible as you can see e.g. with induction. Thank you @Bill!