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.

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.

 A: See the paper Fibonacci Numbers and Identities by Lang and Lang. The authors discuss an elegant framework, involving what they call $\mathcal F$-functions, that unifies proofs for many identities for Fibonacci and Lucas numbers, including d'Ocagne.
A: Call the left-hand side $u_{m,\,n}$ so$$\begin{align}u_{m+1,\,n+1}+u_{m,\,n}&=F_{m+1}F_{n+2}-F_{m+2}F_{n+1}+F_mF_{n+1}-F_{m+1}F_n\\&=F_{m+1}F_{n+2}-F_{m+1}F_{n+1}-F_{m+1}F_n=0,\end{align}$$so some sequence $a_k$ satisfies $u_{m,\,n}=(-1)^na_{m-n}$. To verify $a_k=F_k$, we only need to check the case $n=0$, i.e. $u_{m,\,0}=F_m$, but that's trivial.
A: d'Ocagne identity comes straightly from the matrix form
$$
\eqalign{
  & \left( {\matrix{
   {F_{\,k + 2} }  \cr 
   {F_{\,k + 1} }  \cr 
 } } \right) = \left( {\matrix{
   1 & 1  \cr 
   1 & 0  \cr 
 } } \right)\left( {\matrix{
   {F_{\,k + 1} }  \cr 
   {F_{\,k} }  \cr 
 } } \right)\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {\matrix{
   {F_{\,m + 1} } & {F_{\,n + 1} }  \cr 
   {F_{\,m} } & {F_{\,n} }  \cr 
 } } \right) = \left( {\matrix{
   1 & 1  \cr 
   1 & 0  \cr 
 } } \right)\left( {\matrix{
   {F_{\,m} } & {F_{\,n} }  \cr 
   {F_{\,m - 1} } & {F_{\,n - 1} }  \cr 
 } } \right) =   \cr 
  &  = \left( {\matrix{
   1 & 1  \cr 
   1 & 0  \cr 
 } } \right)^{\,n} \left( {\matrix{
   {F_{\,m - n + 1} } & {F_{\,1} }  \cr 
   {F_{\,m - n} } & {F_{\,0} }  \cr 
 } } \right) = \left( {\matrix{
   1 & 1  \cr 
   1 & 0  \cr 
 } } \right)^{\,n} \left( {\matrix{
   {F_{\,m - n + 1} } & 1  \cr 
   {F_{\,m - n} } & 0  \cr 
 } } \right) \cr} 
$$
and taking the determinant
A: Define:
$$a:=\frac{1+\sqrt{5}}{2}\;\;\;\;\;\;\text{and}\;\;\;\;\;\;\; b:=\frac{1-\sqrt{5}}{2}$$
Then using this follows:
$$F_{n+i}F_{n+j}-F_{n}F_{n+i+j}$$
$$=\left(\frac{a^{\left(n+i\right)}-b^{\left(n+i\right)}}{\sqrt{5}}\right)\left(\frac{a^{\left(n+j\right)}-b^{\left(n+j\right)}}{\sqrt{5}}\right)-\left(\frac{a^{\left(n\right)}-b^{\left(n\right)}}{\sqrt{5}}\right)\left(\frac{a^{\left(n+i+j\right)}-b^{\left(n+i+j\right)}}{\sqrt{5}}\right)$$$$=\frac{\color{red}{a^{\left(2n+i+j\right)}}-a^{\left(n+i\right)}b^{\left(n+j\right)}-a^{\left(n+j\right)}b^{\left(n+i\right)}+\color{blue}{b^{\left(2n+i+j\right)}}}{5}-\frac{\color{red}{a^{\left(2n+i+j\right)}}-a^{\left(n\right)}b^{\left(n+i+j\right)}-a^{\left(n+i+j\right)}b^{\left(n\right)}+\color{blue}{b^{\left(2n+i+j\right)}}}{5}$$$$=\frac{-a^{\left(n+i\right)}b^{\left(n+j\right)}-a^{\left(n+j\right)}b^{\left(n+i\right)}+a^{\left(n\right)}b^{\left(n+i+j\right)}+a^{\left(n+i+j\right)}b^{\left(n\right)}}{5}$$$$=\frac{a^{n}b^{\left(n+j\right)}\left(b^{i}-a^{i}\right)+b^{n}a^{\left(n+j\right)}\left(a^{i}-b^{i}\right)}{5}$$$$=\frac{\left(a^{i}-b^{i}\right)}{\sqrt{5}}\frac{\left(b^{n}a^{\left(n+j\right)}-a^{n}b^{\left(n+j\right)}\right)}{\sqrt{5}}$$$$=\frac{\left(a^{i}-b^{i}\right)}{\sqrt{5}}\left(a^{n}b^{n}\frac{a^{j}-b^{j}}{\sqrt{5}}\right)$$$$=\bbox[5px,border:2px solid #00A000]{\left(-1\right)^{n}F_{i}F_{j}}$$ 

Now I will show that how the other identities can be derived using Vajda's identity :
Set $i \mapsto n-1$, $j \mapsto k$ and fix $n=-1$ in Vajda's identity to get:
$$F_nF_{k+1} -F_1F_{n+k}=(-1)^{1}F_{n-1}F_{k}$$
$$F_nF_{k+1}-(-1)^{1}F_{n-1}F_{k}=F_1F_{n+k}$$
$$F_n\color{red}{F_{k+1}}+F_{n-1}F_{k}=F_1F_{n+k}$$
$$F_n(\color{red}{F_{k}+F_{k-1}})+F_{n-1}F_{k}=F_{n+k}$$
$$F_{k-1}F_n+F_k(F_n+F_{n-1})=F_{n+k}$$
$$\bbox[5px,border:2px solid #00A000]{F_{k-1}F_n+F_kF_{n+1}=F_{n+k}}$$
Which is Honsberger identity.

Set $i \mapsto m-n$ and fix $j = 1$ in Vajda's identity to get:
$$F_mF{n+1}-F_nF_{m+1}=(-1)^{n}F_{m-n}F_1$$
$$\bbox[5px,border:2px solid #00A000]{F_mF_{n+1}-F_{m+1}F_n=(-1)^{n}F_{m-n}}$$
Which is d'Ocagne's identity.

Set $i,j \mapsto r$ and $n \mapsto n-r$ in Vajda's identity to get:
$$F_nF_n-F_{n-r}F_{n+r}=(-1)^{n-r}F_rF_r$$
$$\bbox[5px,border:2px solid #00A000]{F_n^2-F_{n-r}F_{n+r}=(-1)^{n-r}F_r^2}$$
Which is Catalan's identity.

Set $i,j \mapsto r$ , $n \mapsto n-r$ and fix $r = 1$ in Vajda's identity to get:
$$F_nF_n-F_{n-1}F_{n+1}=(-1)^{n-1}F_1F_1$$
$$F_n^2-F_{n-1}F_{n+1}=(-1)^{n-1}F_1^2$$
$$\bbox[5px,border:2px solid #00A000]{F_{n-1}F_{n+1}-F_n^2=(-1)^{n}}$$
Which is Cassini's identity.
