The wikipedia page for Lucas Numbers seems to suggest that if $F_n ≥ 5$ is a Fibonacci number then no Lucas number is divisible by $F_n$. Here is the link.
However, the page does not give any sources cited for me to find a proof of this; I was wondering if somebody could prove this or direct me towards a paper with this theorem.
Suppose $F_n\big| L_k$ where $F_n\ge 5$ (so $n\ge5$) and $k$ is chosen to be minimal.
$\textbf{1)}$ If $k>n$, letting $m=k-n$ gives $L_k=L_{m+n}=L_{m+1}F_n+L_mF_{n-1}$, so
$F_n\big| L_k\implies F_n\big| L_mF_{n-1}\implies F_n\big| L_m$ since $(F_n,F_{n-1})=1$; and this gives a contradiction since $m<k$.
$\textbf{2)}$ If $k\le n$, $\;F_n\big| L_k\implies F_n\le L_k\implies k\ge n-1$
since $F_n=L_{n-2}+F_{n-4}>L_{n-2}$ for $n\ge 5$.
$\textbf{a)}$ If $k=n-1$, then $F_n\big|L_{n-1}$ and $L_{n-1}=F_{n-2}+F_{n}\implies F_n\big|F_{n-2}$, which gives a contradiction.
$\textbf{b)}$ If $k=n$, then $F_n\big|L_n$ and $L_n=F_n+2F_{n-1}\implies F_n\big|2F_{n-1}\implies F_n\big|2$, which gives a contradiction.