# Fibonacci and Lucas numbers congruence relation?

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.

• Why does m<k imply that Fn∣∣Lm is false? – Gust Jul 16 '15 at 1:29
• @HarryChoi $k$ was chosen to be the smallest integer such that $F_n|L_k$, so $m<k$ and $F_n|L_m$ contradicts this. – user84413 Jul 16 '15 at 14:42