# Prove by induction that $F_{2n}=F_nL_n$

In the following exercise from George E. Andrews' Number Theory, we are given that $F_n$ and $L_n$ represent the $nth$ Fibonacci and Lucas numbers respectfully, and we need to prove by induction (i.e. no invocation of the closed form of either)

$$F_{2n}=F_nL_n \qquad(1)$$

With that as our induction hypothesis, to prove the case $n+1$ we have

$$F_{2n+2}=F_{n+1}L_{n+1} \qquad\Rightarrow\qquad F_{2n}+F_{2n+1}=F_{n+1}L_{n+1}=F_nF_{n+1}+F_{n+1}F_{n+2} \qquad (2)$$

My intuition tells me that one needs to somehow derive an identity relating the Fibonacci terms around $n$ and $2n$. While the induction hypothesis would allow us to substitute for the $F_{2n}$ term, reducing the $F_{2n+1}$ term seems a bit more problematic. Furthermore, I am trying to use only the results from the exercises and theorems given up to this point (pgs. 13-17 in the .pdf, up to question 15: https://ia800305.us.archive.org/1/items/NumberTheory_862/Andrews-NumberTheory.pdf)

I cited the results from problems 8 and 9 in which we have

$$F_1+F_3+F_5+\cdots+F_{2n-1}=F_{2n} \qquad (3)$$ $$F_2+F_4+F_6+\cdots+F_{2n}=F_{2n+1}-1 \qquad (4)$$

Summing the two and then substituting into $(2)$, we get

$$F_{2n}+F_{2n+1}=1+F_1+F_2+F_3+\cdots+F_{2n-1}+F_{2n} \qquad\Rightarrow$$ $$1+F_1+F_2+F_3+\cdots+F_{2n-1}+F_{2n}=F_nF_{n+1}+F_{n+1}F_{n+2} \qquad (5)$$

But this still doesn't solve the problem of relating between neighboring terms of $F_{n}$ and $F_{2n}$. Any guidance in the right direction would be much appreciated.

Using $$F_{n} F_{n+1} = \frac{L_{2n+1} - (-1)^{n}}{5}$$ and $L_{n+2} + L_{n} = 5 F_{n+1}$ then \begin{align} F_{2(n+1)} &= F_{n+1} L_{n+1} = F_{n+1} \, (F_{n+2} + F_{n}) \\ &= F_{n+1} F_{n+2} + F_{n} F_{n+1} \\ &= \frac{1}{5} \, ( L_{2n+3} + (-1)^{n} + L_{2n+1} - (-1)^{n} ) \\ &= \frac{1}{5} \, ( 5 F_{2n+2} ) = F_{2n+2}. \end{align}