While I was reading a book, I found the identity \begin{align*} \sum_{k = 0}^{m} (-1)^{[(m - k)/2]}\mathscr{F}_{m}^{k} F_{n + k}^{m - 1} = 0 , \end{align*} where the $ [x] $ denotes the integer that is smaller than $x$, and the symbol $ \mathscr{F} $ is Fibonacci combination define as \begin{align*} \mathscr{F}_{n}^{k} = \frac{F_{n} F_{n - 1} \cdots F_{n - k + 1}}{F_{k} F_{k - 1} \cdots F_{1}} = \prod_{j = 1}^{k} \frac{F_{n - k + j}}{F_{j}} \end{align*} and $ \mathscr{F}_{n}^{0} = 1 $.
Of course it is wrong, we can let $ m = 1 $ to find it is wrong. Then I ask AI, it tells me the right identity is \begin{align*} \sum_{k = 0}^{m} (-1)^{k(k + 1)/2}\mathscr{F}_{m}^{k} F_{n + k}^{m - 1} = 0 \end{align*} and $ m $ is odd. Let us try to let $ m = 3 $, we have \begin{align*} F_{n}^{2} - 2F_{n + 1}^{2} - 2F_{n + 2}^{2} + F_{n + 3}^{2} = 0 \end{align*} It is right. And let $ m = 5 $, we have \begin{align*} F_{n}^{4} - 5F_{n + 1}^{4} - 15F_{n + 2}^{4} + 15F_{n + 3}^{4} + 5F_{n + 4}^{4} - F_{n + 5}^{4} = 0 \end{align*} I don't know does this is right, but if we let $ n = 0 $, it is right.
So, for any odd $ m $, does we have \begin{align*} \sum_{k = 0}^{m} (-1)^{k(k + 1)/2}\mathscr{F}_{m}^{k} F_{n + k}^{m - 1} = 0 \end{align*} I try to use mathematical induction to prove it, but it looks like difficult because the index of $ \mathscr{F} F $ is changed \begin{align*} \mathscr{F}_{2l - 1}^{k} F_{n + k}^{2l - 2} \to \mathscr{F}_{2l + 1}^{k} F_{n + k}^{2l} \end{align*} Can you help me? Thank you.
