Lucas Numbers $(L_n)^2 = L_{2n} \pm 2$

Question

When I was looking at the Lucas Number Series I noticed the following:

• If $n$ is odd, then $(L(n))^2 = L(2n) - 2 $
• If $n$ is even, then $(L(n))^2 = L(2n) + 2 $

Can anyone provide a proof for why this is always true?

Answer 1

It's a standard fact that $L_n = \left( \frac{1+\sqrt{5}}{2} \right)^n + \left( \frac{1-\sqrt{5}}{2} \right)^n$ (and a similar formula for Fibonacci numbers). This is found by solving the recurrence defining the sequence. So \begin{align*} L_n^2 &= \left( \left( \frac{1+\sqrt{5}}{2} \right)^n + \left( \frac{1-\sqrt{5}}{2} \right)^n \right)^2 \\ &= \left( \frac{1+\sqrt{5}}{2} \right)^{2n} + 2 \left( \frac{1+\sqrt{5}}{2} \right)^n \left( \frac{1-\sqrt{5}}{2} \right)^n + \left( \frac{1-\sqrt{5}}{2} \right)^{2n} \\ &= L_{2n} + 2 \left( \frac{1+\sqrt{5}}{2} \right)^n \left( \frac{1-\sqrt{5}}{2} \right)^n \\ &= L_{2n} + 2 \left( \frac{(1+\sqrt{5})(1-\sqrt{5})}{2 \cdot 2} \right)^n \\ &= L_{2n} + 2 \left( \frac{1-5}{4} \right)^n \\ &= L_{2n} + 2 (-1)^n \text{.} \end{align*} Noticing that $n$ odd makes $2(-1)^n = -2$ and $n$ even makes $2(-1)^n = 2$, finishes the proof.