<s>[INST] Are there any number $n$ such that $a_n = 0 \mod (2n + 1) $ where $a_0 = 1, a_1 = 4, a_{n + 2}=3 a_{n + 1} - a_{n}$? Define the sequence $a_n$ by the following.

$$a_0 = 1, a_1 = 4,$$
$$a_{n + 2}=3 a_{n + 1} - a_{n}$$

$a_n ≠ 0 \mod (2n + 1)$ for $1 \le n \le 10^5 $.
Are there any number $n$ such that $a_n = 0 \mod (2n + 1)$?
 [/INST] HINT: Solving the difference equation (use Z Transform):
$$
a_{n+2} = 3 a_{n+1} - a_n \quad\Rightarrow\quad
a_n = \left(\dfrac{3-\sqrt{5}}{2} \right)^n c_1 + \left(\dfrac{3+\sqrt{5}}{2} \right)^n c_2
$$
By conditions $a_0 = 1$ and $a_1 = 4$:
$$
\begin{align}
c_1 + c_2 &= 1 \\
\left( \dfrac{3}{2} - \dfrac{\sqrt{5}}{2} \right) c_1 + \left( \dfrac{3}{2} + \dfrac{\sqrt{5}}{2} \right) c_2&= 4
\end{align}
$$
$$
\therefore \quad c_1 = \dfrac{1-\sqrt{5}}{2} \quad c_2 = \dfrac{1+\sqrt{5}}{2}
$$
Thus:
$$
\begin{align}
a_n &= \left(\dfrac{3-\sqrt{5}}{2} \right)^n \left(\dfrac{1-\sqrt{5}}{2}\right) + \left(\dfrac{3+\sqrt{5}}{2} \right)^n \left(\dfrac{1+\sqrt{5}}{2}\right) \\
\end{align}
$$
Perhaps this can help you a little bit, since putting you in understanding the structure of the numbers you want to test.
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