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 $n=3$.