# Is this a new Fibonacci Identity? [closed]

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

If $$F_n$$ denotes the nth Fibonacci Number, we have the following identity $$$$F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1}$$$$ where $$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!

## closed as off-topic by user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-PuchtaApr 2 at 12:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – user44191, Stopple, Max Alekseyev, Zhi-Wei Sun, Jan-Christoph Schlage-Puchta
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. – user44191 Apr 1 at 22:47
• 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. – user44191 Apr 1 at 22:52
• 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. – user44191 Apr 1 at 23:25
• 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. – Somos Apr 2 at 2:55
• 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. – Sam Hopkins Apr 2 at 20:36

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 $$2$$.
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:
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.