Vajda's Identity/Formulation 1

Theorem

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

Then:

$F_{n + i} F_{n + j} - F_n F_{n + i + j} = \paren {-1}^n F_i F_j$

Proof

From Honsberger's Identity:

\(\ds F_{n + i}\)

\(=\)

\(\ds F_n F_{i - 1} + F_{n + 1} F_i\)

\(\ds F_{n + j}\)

\(=\)

\(\ds F_n F_{j - 1} + F_{n + 1} F_j\)

\(\ds F_{n + i + j}\)

\(=\)

\(\ds F_{i - 1} F_{n + j} + F_i F_{n + j + 1}\)

Therefore:

\(\ds \)

\(\)

\(\ds F_{n + i} F_{n + j} - F_n F_{n + i + j}\)

\(\ds \)

\(=\)

\(\ds \paren {F_n F_{i - 1} + F_{n + 1} F_i} F_{n + j} - F_n \paren {F_{i - 1} F_{n + j} + F_i F_{n + j + 1} }\)

a priori

\(\ds \)

\(=\)

\(\ds \paren {F_n F_{i - 1} + F_{n + 1} F_i} F_{n + j} - F_n F_{i - 1} F_{n + j} - F_n F_i F_{n + j + 1}\)

\(\ds \)

\(=\)

\(\ds \paren {F_n F_{i - 1} + F_{n + 1} F_i - F_n F_{i - 1} } F_{n + j} - F_n F_i F_{n + j + 1}\)

\(\ds \)

\(=\)

\(\ds \paren {F_{n + 1} F_i} F_{n + j} - F_n F_i F_{n + j + 1}\)

\(\ds \)

\(=\)

\(\ds F_i \paren {F_{n + 1} F_{n + j} - F_n F_{n + j + 1} }\)

\(\ds \)

\(=\)

\(\ds F_i \paren {-1}^{2 n + 1} \paren {F_n F_{n + j + 1} - F_{n + 1} F_{n + j} }\)

\(\ds \)

\(=\)

\(\ds F_i \paren {-1}^{n - j - 1} \paren {\paren {-1}^{n + j} F_n F_{n + j + 1} - \paren {-1}^{n + j} F_{n + 1} F_{n + j} }\)

\(\ds \)

\(=\)

\(\ds F_i \paren {-1}^{n - j - 1} \paren {\paren {-1}^{n + j} F_n F_{n + j + 1} + \paren {-1}^{n + j + 1} F_{n + 1} F_{n + j} }\)

\(\ds \)

\(=\)

\(\ds F_i \paren {-1}^{n - j - 1} F_{\paren {n + 1} - \paren {n + j + 1} }\)

Fibonacci Number in terms of Larger Fibonacci Numbers

\(\ds \)

\(=\)

\(\ds F_i \paren {-1}^{n - j - 1} F_{-j}\)

\(\ds \)

\(=\)

\(\ds F_i \paren {-1}^{n - j - 1} \paren {-1}^{j + 1} F_j\)

Fibonacci Number with Negative Index

\(\ds \)

\(=\)

\(\ds \paren {-1}^n F_i F_j\)

$\blacksquare$

Source of Name

This entry was named for Steven Vajda.