Anyone able to provide a solution?

The characteristic equation to the corresponding recurrence relation is $$x^2-2x-41=0.$$Which has roots $$x_1 = 1+\sqrt{42}; x_2 = 1-\sqrt{42}.$$Now, let $a_n =\alpha {x_1}^n + \beta {x_2}^n.$ Then, using the initial conditions, we obtain $$\alpha = -\beta = \frac{1}{\sqrt{42}}.$$$$\therefore \boxed{a_n = \frac{(1+\sqrt{42})^n - (1-\sqrt{42})^n}{\sqrt{42}}}.$$I don’t know how to proceed after here.

Let p be a prime that does not divide 42 and also suppose $x^2 = 42 \mod p$ for some natural $x$. For $p=2017$, $x=\pm 119 \mod 2017$. We claim $a_{p-1}$ is divisible by $p$. We continue the solution from post #3. Working in $\mathbb{Z}/p\mathbb{Z}$, $$a_{p-1} = \frac{(1+\sqrt{42})^{p-1} - (1-\sqrt{42})^{p-1}}{\sqrt{42}} = \frac{(1+x)^{p-1} - (1-x)^{p-1}}{x} = 0.$$The last step follows from Fermat's little theorem.