# Vajda's Identity/Formulation 2

## Theorem

Let $F_n$ be the $n$th Fibonacci number.

Then:

$F_{n + k} F_{m - k} - F_n F_m = \left({-1}\right)^n F_{m - n - k} F_k$

## Proof

We have:

 $\text {(1)}: \quad$ $\ds$  $\ds \left({x^{n + k} - y^{n + k} }\right) \left({x^{m - k} - y^{m - k} }\right) - \left({x^n - y^n}\right) \left({x^m - y^m}\right)$ $\ds$ $=$ $\ds x^{n + m} + y^{n + m} - x^{m - k} y^{n + k} - x^{n + k} y^{m - k} - x^{n + m} - y^{n + m} + x^n y^m + x^m y^n$ $\ds$ $=$ $\ds x^n y^m + x^m y^n - x^{m - k} y^{n + k} - x^{n + k} y^{m - k}$ $\ds$ $=$ $\ds x^n y^n \left({y^{m - n} + x^{m - n} - x^{m - n - k} y^k - x^k y^{m - n - k} }\right)$ $\ds$ $=$ $\ds x^n y^n \left({x^{m - n - k} \left({x^k - y^k}\right) - y^{m - n - k} \left({x^k - y^k}\right)}\right)$ $\text {(2)}: \quad$ $\ds$ $=$ $\ds \left({x y}\right)^n \left({x^{m - n - k} - y^{m - n - k} }\right) \left({x^k - y^k}\right)$

Now substitute:

 $\ds x$ $=$ $\ds \phi$ $\ds y$ $=$ $\ds \hat \phi$

where:

$\phi$ denotes the golden mean
$\hat \phi = 1 - \phi$

first into $(2)$:

 $\ds$  $\ds \left({\phi \hat \phi}\right)^n \left({\phi^{m - n - k} - \hat \phi^{m - n - k} }\right) \left({\phi^k - \hat \phi^k}\right)$ $\ds$ $=$ $\ds \left({-1}\right)^n \left({\phi^{m - n - k} - \hat \phi^{m - n - k} }\right) \left({\phi^k - \hat \phi^k}\right)$ Reciprocal Form of One Minus Golden Mean: $\hat \phi = -\dfrac 1 \phi$ $\ds$ $=$ $\ds \left({-1}\right)^n F_{m - n - k} F_k \times \left({\sqrt 5}\right)^2$ Euler-Binet Formula

and then into $(1)$:

 $\ds$  $\ds \left({\phi^{n + k} - \hat \phi^{n + k} }\right) \left({\phi^{m - k} - \hat \phi^{m - k} }\right) - \left({\phi^n - \hat \phi^n}\right) \left({\phi^m - \hat \phi^m}\right)$ $\ds$ $=$ $\ds F_{n + k} F_{m - k} - F_n F_m \times \left({\sqrt 5}\right)^2$ Euler-Binet Formula

$(1) = (2)$ and hence the result.

$\blacksquare$

## Source of Name

This entry was named for Steven Vajda.