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I have found the following Fibonacci Identity (and proved it).

If $F_n$ denotes the nth Fibonacci Number, we have the following identity \begin{equation} F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1} \end{equation} where $F_1 = F_2 = 1$, $r \leq n$, $h \leq g$, and $n, g, k \in \mathbb{N}$.

It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few.

I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see what the connections are. Thanks for your patience and input!

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    $\begingroup$ This can be simplified to $F_{a - r}F_{b + k + 1} - F_{b - r} F_{a + k + 1} = (-1)^{a + r + 1} F_{b - a} F_{k + r + 1}$, using the substitution $a = n + h, b = n + g$, reducing to 5 variables instead of 6. $\endgroup$
    – user44191
    Commented Apr 1, 2019 at 22:47
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    $\begingroup$ Sorry, I meant 4 variables instead of 5. I'm pretty sure this can be reduced by changes of variables (with no Fibonacci arithmetic) to Vajda's identity. $\endgroup$
    – user44191
    Commented Apr 1, 2019 at 22:52
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    $\begingroup$ This is a disguised version of Vajda's identity (with minor amounts of arithmetic for powers of $-1$); try doing variable substitutions to see for yourself. $\endgroup$
    – user44191
    Commented Apr 1, 2019 at 23:25
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    $\begingroup$ The identity is with arguments renamed the same as $F_a F_b - F_c F_{a+b-c} = (-1)^{a+1}F_{c-a}F_{b-c}$. Read my essay "In the elliptic realm" to see connection. $\endgroup$
    – Somos
    Commented Apr 2, 2019 at 2:55
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    $\begingroup$ I don't really see why this question should be put on hold. The answer is "no, this identity is not new" but the question itself seems totally legitimate and not obvious. $\endgroup$
    – Sam Hopkins
    Commented Apr 2, 2019 at 20:36

3 Answers 3

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Here is an expanded comment of user44191. The basic observation is that one can extend $F_n$ to all $n\in {\mathbb Z}$ by requiring $F_{-n}=(-1)^{n+1}F_n$. Then by Vajda's formula, one has $$F_{n'+a'}F_{n'+b'}-F_{n'}F_{n'+a'+b'}=(-1)^{n'}F_{a'}F_{b'}=(-1)^{n'+a'+1}F_{-a'}F_{b'},$$ where one uses the extension above in the last equality. Now by the following substitutions, the above identity leads to the one given by user44191: $$n'=b-r,a'=a-b,b'=k+r+1,$$ using the fact that $a-r+1$ is congruent to $a+r+1$ mod $2$.

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This identity is a special case of Euler's Identity for Continuants. It is a Pfaffian of degenerate $4\times 4$ matrix. Concrete mathematics gives the following reference:

enter image description here

As Michael Somos mentioned in his comment it is a part of "elliptic realm" where different identities arise as determinants of degenerate matrices. These matrices are degenerate because they are submatrices of infinite matrices of finite rank. For examle the matrix with entries $a_{m,n}=s_{m+n}s_{m-n}$ $(m,n\in \mathbb{Z})$ where $s_n$ is the Somos-$4$ seqence has rank $2$. For Somos-$6$ corresponding matrix has rank $4$ etc.

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"Vajda's identity" is really Tagiuri's identity: A. Tagiuri, Di alcune successioni ricorrenti a termini interi e positivi, Periodico di Matematica 16 (1900–1901), 1–12. See also https://math.stackexchange.com/questions/1356391/is-there-a-name-for-this-fibonacci-identity.

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  • $\begingroup$ While probably historically interesting, this doesn't seem relevant to the question. $\endgroup$
    – LSpice
    Commented Apr 2, 2019 at 15:19
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    $\begingroup$ The question is "I am still wondering if anyone has seen this identity before?" I am saying, yes, this identity (or an equivalent identity) was seen by Tagiuri in 1900. $\endgroup$
    – Ira Gessel
    Commented Apr 2, 2019 at 16:26

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