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The Fibonacci and Lucasnumbers are defined for all integers $n$ by the recurrence relations $$F_n=F_{n-1}+F_{n-2}\text{ where }F_1=1\text{ and }F_2=1,$$ $$L_n=L_{n-1}+L_{n-2}\text{ where }L_1=1\text{ and }L_2=3.$$ I would like to find the cleanest proof that $$F_nF_{n+1}=\frac{1}{4}(F_{n+2}^2-F_{n-1}^2).$$ Note: I can prove this result using induction, so that is not of interest to me. I can also prove it using Binet's formula, which is obviously doesn't fulfil my want for a "clean" proof. I am aware of formulae such as \begin{align*} F_{n+k}+F_{n-k}=F_nL_k\text{ where $k$ is even,}\\ F_{n+k}+F_{n-k}=L_nF_k\text{ where $k$ is odd,}\\ F_{n+k}-F_{n-k}=F_nL_k\text{ where $k$ is odd,}\\ F_{n+k}-F_{n-k}=L_nF_k\text{ where $k$ is even.}\\ \end{align*} but these don't seem to be of help as the different between $n+2$ and $n-1$ is $3$ (i.e. the difference is odd). Does anyone have any suggestions for a one line proof (with the use of a suitable identity)?

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The "cleanest" proof is not necessarily a one-line proof. That said, $$\begin{align*} F_{n+2}^2 - F_{n-1}^2 &= (F_{n+2} + F_{n-1})(F_{n+2} - F_{n-1}) \\ &= ((F_{n+1} + F_n) + (F_{n+1} - F_n))((F_{n+1} + F_n) - (F_{n+1} - F_n)) \\ &= (2F_{n+1})(2F_n) \\ &= 4F_{n+1}F_n, \end{align*}$$ and the result immediately follows.

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  • $\begingroup$ You're right. This is the nicest proof. I was kind of looking for a technique that might find a nice expression for $F_n^2-F_m^2$ in the instance where $n-m$ is odd. I should ask this as a new question though. $\endgroup$ – Auslander Mar 17 '16 at 7:55
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We have \begin{align} F_{n+2}^2 &= (F_n+F_{n+1})^2 = F_n^2 + F_{n+1}^2 + 2F_nF_{n+1} = 2F_n F_{n+1} + (F_n+F_{n-1})^2 + F_n^2\\ & = 2F_nF_{n+1} + 2F_n^2 + 2F_n F_{n-1} + F_{n-1}^2 = 2F_nF_{n+1} + 2F_n(F_n+F_{n-1}) + F_{n-1}^2\\ F_{n+2}^2 - F_{n-1}^2 & = 4F_nF_{n+1} \end{align}

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  • $\begingroup$ Thanks. What I am wondering is: can it be done in one line with a suitable identity? $\endgroup$ – Auslander Mar 17 '16 at 4:40
  • $\begingroup$ I've adjusted the question to reflect this desire! $\endgroup$ – Auslander Mar 17 '16 at 4:45

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