Proof of d'Ocagne's identity

It's known that Fibonacci numbers satisfy the following relation:

$$F_mF_{n+1}-F_{m+1}F_n=(-1)^nF_{m-n}$$

Which is called d'Ocagne's identity.

This identity with the following identities are well-known:

$$F_{n-1}F_{n+1}-F_{n}^2=(-1)^n\tag{Cassini's identity}$$ $$F_{n}^2-F_{n-r}F_{n+r}=(-1)^{n-r}F_r^2\tag{Catalan's identity }$$ $$F_{n+i}F_{n+j}-F_{n}F_{n+i+j}=(-1)^{n}F_iF_j\tag{Vajda's identity }$$ $$F_{k−1}F_n + F_kF_{n+1} = F_{n+k} \tag{Honsberger identity}$$

Cassini's identity is a special case of Catalan's identity and can be derived with $r=1$.

The usual way for proving these identities is using $2×2$ matrix , another way would be induction,I know how to prove Catalan's identity using induction but still I have not seen any proof of d'Ocagne's identity,I'm asking if someone know a proof of that (induction preferred)?

Also is their any combinatorial poof for d'Ocagne's identity? if yes, so it would be really nice to see the proof.

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My try:

• Define: $$a:=\frac{1+\sqrt{5}}{2}\;\;\;\;\;\;\text{and}\;\;\;\;\;\;\; b:=\frac{1-\sqrt{5}}{2}$$ Then using this follows: $$F_mF_{n+1}-F_{m+1}F_n$$ $$=\left(\frac{a^{m}-b^{m}}{\sqrt{5}}\right)\left(\frac{a^{\left(n+1\right)}-b^{\left(n+1\right)}}{\sqrt{5}}\right)-\left(\frac{a^{\left(m+1\right)}-b^{\left(m+1\right)}}{\sqrt{5}}\right)\left(\frac{a^{n}-b^{n}}{\sqrt{5}}\right)$$

$$=\frac{\color{red}{a^{\left(m+n+1\right)}}-a^{m}b^{\left(n+1\right)}-a^{\left(n+1\right)}b^{m}+\color{blue}{b^{\left(m+n+1\right)}}}{5}-\frac{\color{red}{a^{\left(m+n+1\right)}}-a^{\left(m+1\right)}b^{n}-a^{n}b^{\left(m+1\right)}+\color{blue}{b^{\left(m+n+1\right)}}}{5}$$ $$=\frac{-a^{m}b^{\left(n+1\right)}-a^{\left(n+1\right)}b^{m}+a^{\left(m+1\right)}b^{n}+a^{n}b^{\left(m+1\right)}}{5}$$$$=\frac{a^{m}b^{n}\left(a-b\right)+a^{n}b^{m}\left(b-a\right)}{5}=\frac{\left(a-b\right)\left(a^{m}b^{n}-a^{n}b^{m}\right)}{5}$$$$=\left(a-b\right)\frac{\left(a^{\left(m-n\right)}-b^{\left(m-n\right)}\right)}{\sqrt{5}}\frac{a^{n}b^{n}}{\sqrt{5}}$$$$=\left(a-b\right)\frac{a^{n}b^{n}}{\sqrt{5}}F_{m-n}$$$$=\bbox[5px,border:2px solid #00A000]{\left(-1\right)^{n}F_{m-n}}$$

Which is the claim.

• Another way for proving this identity would be setting $i \mapsto m-n$ and $j \mapsto 1$ in Vajda's identity.

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