A sequence $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and find $x_{2020}$ The sequence is given by the formula $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$ and it is known that  $x_{2017} + x_{2023} = 990$, then what is $x_{2020}$ = ?
My little approch:
It is given that  $x_{2017} + x_{2023} = 990$ ----- (1)  and $x_{n+1} = 3x_n + \sqrt{8x^2_n + 2}$
So,
$x_{2017} = x_{2016+1} = 3 x_{2016} + \sqrt{8x^2_{2016} + 2}$ -------(2)
$x_{2023} = x_{2022+1} = 3 x_{2022} + \sqrt{8x^2_{2022} + 2}$ -------(3)
Then from (1),(2),(3) =>
$3 x_{2016} + \sqrt{8x^2_{2016} + 2} + 3 x_{2022} + \sqrt{8x^2_{2022} + 2} = 990$
$3 (x_{2016}+x_{2022} ) + \sqrt{8x^2_{2016} + 2}  + \sqrt{8x^2_{2022} + 2} = 990$
$3 (x_{2016}+x_{2020} + x_{2021}) + \sqrt{8x^2_{2016} + 2}  + \sqrt{8x^2_{2020} + 2}+ \sqrt{8x^2_{2021} + 2}+ \sqrt{8x^2_{2022} + 2} = 990$
I stuck here and can't go further. Please help me with it.

After you get $x_{n+1} - 6x_n + x_{n-1}=0$ as in the other two answers, a conceptually  easy way to establish the relationship between $x_n$ and $x_{n-3}+x_{n+3}$ is as follows:
Rewrite the recurrence equation as
$$(\mathbb E^2-6\mathbb E+1)x_n=0$$
where $\mathbb E$ is the forward shift operator $\mathbb E^{i} x_n = x_{n+i}, \forall i \in \mathbb N$. Then the characteristic equation is $$\lambda^2 - 6\lambda + 1 =0$$ with two roots $a, b$ such that $a+b=6, ab=1$. Now
$$a^3\cdot b^3 =1, a^3+b^3 = (a+b)((a+b)^2-3ab)=6\cdot (6^2-3)=198$$
Therefore $$\lambda^6 - 198 \lambda^3+1=(\lambda^3-a^3)(\lambda^3-b^3)=g(\lambda)(\lambda-a)(\lambda-b)=g(\lambda)(\lambda^2-6\lambda+1)$$
where $g(\lambda)=(\lambda^2 + a\lambda +a^2)(\lambda^2 + b\lambda +b^2)$ (we don't need to know the coefficients, we only need the fact that $g$ is a polynomial)
Then
$$(\mathbb E^6 - 198 \mathbb E^3 + 1) x_n = g(\mathbb E)(\mathbb E^2 - 6\mathbb E+1)x_n = 0\\
\implies x_{n+6} - 198 x_{n+3} + x_n=0 \implies x_{2020} = 990/198=5.$$
