Cassini's Identity

Theorem
Let $F_k$ be the $k$th Fibonacci number.

Then $F_{n+1}F_{n-1} - F_n^2 = \left({-1}\right)^n$.

This is also sometimes reported (slightly less elegantly) as $F_{n+1}^2 - F_n F_{n+2} = \left({-1}\right)^n$

Proof
We see that $F_2 F_0 - F_1^2 = 1 \times 0 - 1 = -1 = \left({-1}\right)^1$, so the proposition holds for $n=1$.

We also see that $F_3 F_1 - F_2^2 = 2 \times 1 - 1 = \left({-1}\right)^2$, so the proposition holds for $n=2$.

Suppose the proposition is true for $n=k$, that is, $F_{k+1}F_{k-1} - F_k^2 = \left({-1}\right)^k$.

We now see whether we can show that it follows from this that the proposition is true for $n=k+1$, that is, $F_{k+2}F_k - F_{k+1}^2 = \left({-1}\right)^{k+1}$.

So:



\begin{align} F_{k+2} F_k - F_{k+1}^2 & = \left({F_k + F_{k+1}}\right) F_k - F_{k+1}^2 \\ & = F_k^2 + F_k F_{k+1} - F_{k+1}^2 \\ & = F_k^2 + F_k F_{k+1} - F_{k+1} \left({F_k + F_{k-1}}\right) \\ & = F_k^2 + F_k F_{k+1} - F_k F_{k+1} - F_{k+1} F_{k-1} \\ & = F_k^2 - F_{k+1} F_{k-1} \\ & = \left({-1}\right) \left({F_{k+1} F_{k-1} - F_k^2}\right) \\ & = \left({-1}\right) \left({-1}\right)^k \\ & = \left({-1}\right)^{k+1} \\ \end{align} $

So by the principle of mathematical induction, the proof is complete.

Note that from the above we have that $F_{k+2} F_k - F_{k+1}^2 = \left({-1}\right)^{k+1}$ from which $F_{n+1}^2 - F_n F_{n+2} = \left({-1}\right)^n$ follows immediately.

Proof using Matrices
The same thing can be proved more elegantly by starting with the proof by induction of this identity:



\begin{bmatrix} F_{n+1} & F_n     \\ F_n     & F_{n-1} \end{bmatrix} = \begin{bmatrix} 1    & 1    \\  1     & 0 \end{bmatrix}^n $

and then taking the determinant of both sides.

Base case:



\begin{bmatrix} F_{2} & F_1     \\ F_1     & F_{0} \end{bmatrix} = \begin{bmatrix} 1    & 1    \\  1     & 0 \end{bmatrix} = \begin{bmatrix} 1    & 1    \\  1     & 0 \end{bmatrix}^1 $

Induction hypothesis:



\begin{bmatrix} F_{k+1} & F_k     \\ F_k     & F_{k-1} \end{bmatrix} = \begin{bmatrix} 1    & 1    \\  1     & 0 \end{bmatrix}^k $

The induction step follows from conventional matrix multiplication:



\begin{bmatrix} F_{k+1} & F_k     \\ F_k     & F_{k-1} \end{bmatrix} \begin{bmatrix} 1    & 1    \\  1     & 0 \end{bmatrix} = \begin{bmatrix} F_{k+1}+ F_k & F_{k+1}     \\ F_{k+1}     & F_{k} \end{bmatrix} = \begin{bmatrix} F_{k+2} & F_{k+1}     \\ F_{k+1} & F_{k} \end{bmatrix} $

So by induction:

\begin{bmatrix} F_{n+1} & F_n     \\ F_n     & F_{n-1} \end{bmatrix} = \begin{bmatrix} 1    & 1    \\  1     & 0 \end{bmatrix}^n $

Now we calculate the determinants:

The LHS follows directly from the order 2 determinant:


 * $\begin{bmatrix}

F_{n+1} & F_n     \\ F_n     & F_{n-1} \end{bmatrix} = F_{n+1} F_{n-1} - F_n^2$

Now for the RHS:

Base case:

\begin{vmatrix} 1    & 1    \\  1     & 0 \end{vmatrix} = 1 \times 0 - 1 \times 1 = -1 = \left({-1}\right)^1 $

Induction hypothesis:

\begin{vmatrix} 1    & 1    \\  1     & 0 \end{vmatrix}^k = \left({-1}\right)^k $

The induction step follows from Determinant of Matrix Product:



\begin{vmatrix} 1    & 1    \\  1     & 0 \end{vmatrix}^{k+1} = \begin{vmatrix} 1    & 1    \\  1     & 0 \end{vmatrix}^k \begin{vmatrix} 1    & 1    \\  1     & 0 \end{vmatrix} = \left({-1}\right)^k \left({-1}\right) = \left({-1}\right)^{k+1} $

Hence the result by induction.

He first published it in 1680.

However, it had been mentioned as early as 1608 in a letter by Johannes Kepler.