Relationship between Fibonacci and Lucas numbers

Đầu tiên, ta nhắc lại các định nghĩa của dãy Fibonacci và dãy Lucas như sau:
a) Dãy Fibonacci: $F_0=0;F_1=1;F_{n+2}=F_{n+1}+F_{n},\forall n\ge 2,n\in \mathbb{N}$.
b) Dãy Lucas: $L_0=2;L_1=1;L_{n+2}=L_{n+1}+L_{n},\forall n\ge 2,n\in \mathbb{N}$.
Và dưới đây là mối liên hệ giữa dãy Fibonacci và dãy Lucas.
1) $L_{2n}=L_{n}^2+2.(-1)^{n+1}$.
2) $L_{2n}.L_{2n+2}-5F_{n+1}^2=1$.
3) $\sum\limits_{k=1}^{n}F_{2k}=F_{2n+1}-1$
4) $\sum\limits_{k=1}^{n}F_{2k-1}=F_{2n}$
5) $\sum\limits_{k=1}^{n}(-1)^{k+1}F_k=(-1)^{n+1}F_{n-1}+1$
6) $L_n=F_{n+1}+F_{n-1}$
7) $5F_n=L_{n+1}+L_{n-1}$
8) $F_{n+m}=F_{n-1}F_{m}+F_{n}F_{m+1}$
9) $F_{n+m}=\frac{L_nF_m+L_mF_n}{2}$
10) $L_{n+m}=F_mL_{n+1}+F_{m-1}L_n$
11) $F_{n-m}=(-1)^m.\frac{L_mF_n-F_mL_n}{2}$
12) $L_{n-m}=(-1)^m.\frac{L_mL_n-5F_mF_n}{2}$
13) $F_nF_m-F_{n-k}F_{m+k}=(-1)^{n-k}.F_k.F_{m+k-n}$
14) $F_{n+3}^2=2F_{n+2}^2+2F_{n+1}^2-F_n^2$
15) $F_{n+4}^3=3F_{n+3}^3+6F_{n+2}^3-3F_{n+1}^3-F_n^3$
Source: https://artofproblemsolving.com/community/c260h1565488p9589231

Lucal sequences

Problem: Let $(L_n)$ be the Lucas sequence, that is, $L_0=2,L_1=1,L_{n+2}=L_{n+1}+L_n$. Prove that: $L_n$ is a perfect square if and only if $n=3$.
Solution:
If $n$ is even, we have $L_{2m}=L_{m}^2-2(-1)^{n-1}$ which gives a contradiction.
If $n$ is odd
Case 1: $n=4t+1$ and $n\ne 1$. We write $n=1+2.3^r.k$ so $k$ is not divisible by $3$ and $2|k$. We have $L_{m+2k}\equiv -L_m(\text{ mod }L_k)$ so $L_n\equiv -L_1\equiv -1(\text{ mod }L_k)$.
And for every even $k$ satisfy $k$ is not divisible by $3$, we can check that $L_k\equiv 3(\text{ mod }4)$ so if $L_n=x^2$, we have $x^2+1$ has a prime divisor with the form $4k+3$.
Case 2: $n=4t+3$, we write $n=3+2.3^r.k$, then do similar with the first case we obtain $x^2+4$ has a prime divisor with the form $4k+3$ with $x$ is odd. So $x$ is even, then $L_n=4$ which gives $n=3$.