Logical Algebra (Problem 2 - Version 2)

The Lucas numbers are related to the Fibonnaci numbers. Their relation identities are listed in the picture below: Also, a closed-form expression of the Lucas numbers is given below: And finally, the closed-form expression of the Lucas numbers being combined with the Binet formula to create a formula for $[{\displaystyle \phi }]^n$ : So, my question to you is this: how many of the Fibonnaci - Lucas numbers relation identities are equivalent to each other? (-1 if none, >= 1 if any of the identities are equivalent to each other.)

Bonus: Prove that the addition of the first 7 Lucas polynomials is equivalent to: $x^3 + (x + 6) (x + 1)$ $+$ $\frac{14x^2 + 9x + 8}{x^3}$ . Here are the first seven Lucas polynomials: $2$ , $x + 2$ , $x^2 + 2$ , $x^3 + 3x$ , $x^4 + 4x^2 + 2$ , $x^5 + 5x^3 + 5x$ and $x^6 + 6x^4 + 9x^2 + 2$ (this is Logical Algebra (Problem 3) - how to get there explained in Logical Algebra (Problem 3).)

Bonus 2: Prove that $F_3(x) + 2F_2(x) = (x + 1) ^2$ where $F_n(x)$ is a Fibonnaci polynomial.

Bonus 3: Prove that $L_1(x) + F_2(x) + L_0(x) = 2 (x + 1)$ where $F_n(x)$ is a Fibonnaci polynomial and $L_n(x)$ is a Lucas polynomial.

Challenge: $\frac{n(n + 1)}{2}$ = $L_n(19)$ where $L_n(19)$ is the 19th Lucas number. Using $L_n = [{\displaystyle \phi^n}]$ , find the 19th Lucas number and hence prove using $\frac{n(n + 1)}{2}$ that the 19th Lucas number is one of only three triangular numbers in the Lucas series (or the Lucas number sequence.)

Place your answers to Bonuses 2, 3 and the Challenge in the Discussion Section.