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- #1

- Apr 13, 2013

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I want to show that for each $n \geq 1$ it holds that $2^n L_n \equiv 2 \pmod{10}$.

$L_n$ is the Lucas sequence.

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.

Could you give me a hint how we get the desired congruence?