# Fibonacci conjecture: $(F_{n+5})^2 - (F_n)^2 = 3((F_{n+3})^2 - (F_{n+2})^2) + 8 F_{n+2} F_{n+3}$.

So this is the question I have

The Fibonacci sequence is a recurrence system given by $$F_1 = 1, \ F_2 = 1, \ F_{n+2} = F_{n+1} + F_n \qquad (n = 1, 2, 3, \ldots).$$ This question concerns the following conjecture: $$(F_{n+5})^2 - (F_n)^2 = 3((F_{n+3})^2 - (F_{n+2})^2) + 8 F_{n+2} F_{n+3} \qquad (n = 1, 2, 3, \ldots).$$ (a) Confirm that the conjecture is true when $n = 6$.

(b) Prove that the conjecture is true for all integers $n \geq 1$.

But when I go through the solution I don't understand one line (separated and marked).

(b) Let $n$ be any integer such that $n \geq 1$.

\begin{align} \left( F_{n+5} \right)^2 - \left( F_n \right)^2 &= \left( F_{n+5} - F_n \right) \left( F_{n+5} + F_n \right) \\ &= \left( F_{n+4} + F_{n+3} - F_{n+2} + F_{n+1} \right) \left( F_{n+4} + F_{n+3} + F_{n+2} - F_{n+1} \right) \\ \\ &= \left( \color{red}{2} F_{n+3} + F_{n+1} \right) \left( 2 F_{n+3} + 2 F_{n+2} - F_{n+1} \right) \tag{*} \\ \\ &= \left( 2 F_{n+3} - F_{n+3} - F_{n+2} \right) \left( 2 F_{n+3} + 2 F_{n+2} - F_{n+3} + F_{n+2} \right) \\ &= \left( 3 F_{n+3} - F_{n+2} \right) \left( F_{n+3} + 3 F_{n+2} \right) \\ &= 3 \left( \left( F_{n+3} \right)^2 - \left( F_{n+2} \right)^2 \right) + 8 F_{n+2} F_{n+3} \end{align} Hence, the conjecture is true for all integers $n \geq 1$

In the marked line, where does the red $\color{red}{2}$ come from?

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I replaced the solution image with $\LaTeX$. Feel free to revert if you believe something was lost in transcription. – arjafi Nov 11 '12 at 14:50
@ArthurFischer nothing is lost..tks for doing that! – JackyBoi Nov 11 '12 at 14:59

He substituted $F_{n+3}+F_{n+2}$ for $F_{n+4}$.
Use $F_{n+4}-F_{n+2}=F_{n+3}$ in the first factor and $F_{n+4}=F_{n+3}+F_{n+2}$ in the second factor to obtain the asterisked line from the preceding one.