<s>[INST] Congruence with Lucas sequence I want to show that for each $n \geq 2$ the following congruence holds,
$$L_{2^n} \equiv 7 \pmod{10}.$$
According to my notes,
$$L_n=\left( \frac{1+\sqrt{5}}{2}\right)^n+\left( \frac{1-\sqrt{5}}{2}\right)^n$$
and
$$L_n=F_{n-1}+F_{n+1},$$
where $F_n$ is the $n$-th Fibonacci number.
Do we use somehow the last equality in order to show the desired congruence?
 [/INST] The following helps :
$$L_{2n}=L_n^2-2(-1)^n\tag1$$
(the proof is written at the end of the answer.)

This is a proof by induction.
For $n=2$, $L_{2^2}=7\equiv 7\pmod{10}.$
Suppose that $L_{2^n}=10k+7$ for some $k\in\mathbb N$.
Then, using $(1)$, we get
$$\begin{align}L_{2^{n+1}}&=L_{2^n}^2-2(-1)^{2^n}
\\\\&=(10k+7)^2-2
\\\\&=10(10k^2+14k+4)+7
\\\\&\equiv 7\pmod{10}\end{align}$$

Proof for $(1)$ :
Using that $L_{n}=\alpha^n+\beta^n$ where $\alpha=\frac{1-\sqrt 5}{2}$ and $\beta=\frac{1+\sqrt 5}{2}$, we get
$$L_{2n}-L_n^2=\alpha^{2n}+\beta^{2n}-(\alpha^n+\beta^n)^2=-2(\alpha\beta)^n=-2(-1)^n$$
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