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 $a_n = 0 \mod (2n + 1)$?
No, there are no others. You have the original Lucas numbers of odd index. The device of using $2n+1$ means they are asking about the Lucas numbers divisible by their index, see https://oeis.org/A016089  It seems this is proved in the article Noam mentions, (in any case it is in the Examples) with easier to read in Somer. The two books with articles by Somer were edited by Gerald E. Bergum, Andreas N. Philippou, and A. F. Horadam. 

There are plenty of even indexes that divide the Lucas number, but the only odd one is $1$ itself. 

Note that $L_{n+2} = L_{n+1} + L_n$ gives $L_{n+1} = L_{n+2} - L_n,$
$$ L_{n+3} = L_{n+2} + L_{n+1} = L_{n+2} +  L_{n+2} - L_n = 2 L_{n+2} - L_n , $$
$$ L_{n+4} = L_{n+3} + L_{n+2} = 2 L_{n+2} - L_n +L_{n+2} = 3 L_{n+2} - L_n.  $$
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 $a_1 = 4$:
$$
\begin{align}
c_1 + c_2 &amp;= 1 \\
\left( \dfrac{3}{2} - \dfrac{\sqrt{5}}{2} \right) c_1 + \left( \dfrac{3}{2} + \dfrac{\sqrt{5}}{2} \right) c_2&amp;= 4
\end{align}
$$
$$
\therefore \quad c_1 = \dfrac{1-\sqrt{5}}{2} \quad c_2 = \dfrac{1+\sqrt{5}}{2}
$$
Thus:
$$
\begin{align}
a_n &amp;= \left(\dfrac{3-\sqrt{5}}{2} \right)^n \left(\dfrac{1-\sqrt{5}}{2}\right) + \left(\dfrac{3+\sqrt{5}}{2} \right)^n \left(\dfrac{1+\sqrt{5}}{2}\right) \\
\end{align}
$$
Perhaps this can help you a little bit, since putting you in understanding the structure of the numbers you want to test.