I'm trying to prove the following: $$L_k^2-5F_k^2=4(-1)^k\qquad k\ge1$$
$L_k$ is the $k$th term of the Lucas numbers and $F_k$ is the $k$th term of the Fibonacci sequence.
I've tried using mathematical induction, however it's not working out too well. I tried starting out by manipulating $L^2_{k+1}-5F^2_{k+1}$, but I can't prove that it equals $4(-1)^{k+1}$.
Any help is greatly appreciated!
On
In this answer, it is shown that $$ F_nF_k+F_{n-1}F_{k-1}=F_{n+k-1}\tag1 $$ The Lucas Numbers satisfy the same recurrence as the Fibonacci Numbers: $$ L_n=L_{n-1}+L_{n-2}\tag2 $$ Therefore, since $$ \color{#C00}{L_1}L_1+\color{#C00}{L_0}L_0=5=\color{#C00}{5F_1}\quad\text{and}\quad\color{#C00}{L_2}L_1+\color{#C00}{L_1}L_0=5=\color{#C00}{5F_2}\tag3 $$ equations $(3)$ and recurrence $(2)$ inductively imply $$ \color{#C00}{L_n}L_1+\color{#C00}{L_{n-1}}L_0=\color{#C00}{5F_n}\tag4 $$ Furthermore, since $$ \color{#C00}{L_1}L_2+\color{#C00}{L_0}L_1=5=\color{#C00}{5F_2}\quad\text{and}\quad\color{#C00}{L_2}L_2+\color{#C00}{L_1}L_1=10=\color{#C00}{5F_3}\tag5 $$ equations $(5)$ and recurrence $(2)$ inductively imply $$ \color{#C00}{L_n}L_2+\color{#C00}{L_{n-1}}L_1=\color{#C00}{5F_{n+1}}\tag6 $$ Equations $(4)$ and $(6)$ and recurrence $(2)$ inductively imply $$ L_nL_k+L_{n-1}L_{k-1}=5F_{n+k-1}\tag7 $$ Setting $k=n$ in $(1)$ and $(7)$ gives $$ F_n^2+F_{n-1}^2=F_{2n-1}\tag8 $$ and $$ L_n^2+L_{n-1}^2=5F_{2n-1}\tag9 $$ Subtracting $5$ times $(8)$ from $(9)$ and rearranging yields $$ \begin{align} L_n^2-5F_n^2 &=-\left(L_{n-1}^2-5F_{n-1}^2\right)\\ &=(-1)^n\left(L_0^2-5F_0^2\right)\\[2pt] &=(-1)^n\,4\tag{10} \end{align} $$
We may use Binet's formulas: $$L_k=\varphi^k+(-1/\varphi)^k\quad\text{and}\quad \sqrt{5}F_k=\varphi^k-(-1/\varphi)^k$$ where $\varphi=(1+\sqrt{5})/2$. Then, after factoring the difference of squares, we get $$L_k^2-5F_k^2=(\varphi^k+(-1/\varphi)^k)^2-(\varphi^k-(-1/\varphi)^k)^2 =(2\varphi^k)\cdot(2(-1/\varphi)^k)=4(-1)^k.$$