<s>[INST] Is every $N$th Fibonacci number where $N$ is divisible by $5$ itself divisible by $5$ There seems to be a pattern of sorts in the Fibonacci sequence:
The $5$th, $10$th, $15$th & $20$th values are:
$$5, 55, 610, 6765$$
Does this pattern continue ad infinitum? I've tried a few more using Binet's Formula and it seems to hold. So:

Is every $N$th Fibonacci number where $N$ is divisible by $5$ itself divisible by $5$?
Bonus Q: Are there any other patterns?

 [/INST] Let us make a table of the Fibonacci sequence modulo $5$. If we can find two occurrences of the same two terms modulo $5$ with all the $F_{5k}$ (between those two occurence) being $0$ modulo $5$, we can prove this statement.
$$\begin{array}{c|c|c|} 
\text{$F_1$ to $F_5$} & \text{1} & \text{1} & \text{2} & \text{3} & \text{0} \\ 
\text{$F_6$ to $F_{10}$} & \text{3} & \text{3} & \text{1} & \text{4} & \text{0} \\ 
\text{$F_{11}$ to $F_{15}$} & \text{4} & \text{4} & \text{3} & \text{2} & \text{0}  \\ 
\text{$F_{16}$ to $F_{20}$} & \text{2} & \text{2} & \text{4} & \text{1} & \text{0} \\ 
\text{$F_{21}$ to $F_{25}$} & \text{1} & \text{1} & \text{2} & \text{3} & \text{0} \\ 
\end{array}$$
Since $F_1$ and $F_2$ are the same as $F_{21}$ and $F_{22}$, and $F_5$,  $F_{10}$, $F_{15}$ and $F_{20}$ are all $0$, then this cycle repeats indefinitely, and hence $F_{5k} \equiv 0 \pmod 5$.
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