I am trying to proof read a paper from Zhi Hong Sun, however I cannot reconcile several of the results. I may have made an error, so I would like a confirmation from the community here.
Does the following theorem work for all positive integers $k$ and $n$?
Let $k$ and $n$ be integers with $k\ne 0$. Let $F_n$ be a Fibonacci number, and let $L_n$ be a Lucas number. Then \begin{equation}F_{kn}/F_k \equiv \begin{cases} (-1)^{km}(2m+1) &\pmod{5F_k^{2}} & \quad \text{if } n=2m +1,\\ (-1)^{k(m-1)}mL_k &\pmod{5F_k^{2}} & \quad \text{if } n=2m. \end{cases} \end{equation}
Yes, the congruence hold for all integers $k,n\in\mathbb{Z}$ with $k\not=0$. Note that $$F_{k+j}=L_kF_j-(-1)^{k}F_{j-k}$$ (use induction with respect to $j$ and recall that $F_{-n}=(-1)^{n-1}F_n$). Hence, by letting $j=nk$, we have that $$F_{(n+1)k}=F_{k+nk}=L_kF_{nk}-(-1)^{k}F_{(n-1)k}.$$ Now we can verify the congruence by induction with respect to $n$.
P.S. For $n=12$ and $k=3$, then $F_{kn}/F_k=7465176$, $ (-1)^{k(m-1)}(mL_k)=-24$ and $7465176$ and $-24$ are congruent to $16$ modulo $5F_k^{2}=20$.