multiply two Fibonacci numbers

Question

$0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots$

As I noticed if you multiply $T_{n-1}$ and $T_{n+1}$ from Fibonacci numbers, you almost always get $(T_{n})^2 \pm 1$. for example:

$2 \times 5 = 3^2 + 1$

$3 \times 8 = 5^2 - 1$

$5 \times 13 = 8^2 + 1$

$13 \times 34 = 21^2 + 1$

So I was wondering, is there any algebraic proof for this?

https://en.wikipedia.org/wiki/Cassini_and_Catalan_identities, https://math.stackexchange.com/questions/2415088/proof-cassinis-identity-with-induction-and-fibonacci-sequence — Minus One-Twelfth, Mar 01 '21 at 09:24

score 2 · Accepted Answer · answered Mar 01 '21 at 09:24

Yes. A way to see this identity:

Let $Q = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$. Then, $Q^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}$, where $F_i$ is the $i$-th Fibonacci number (a fun exercise would be to check this - it's a fairly straightforward induction argument).

Then, we notice $\det(Q) = -1$, so $\det(Q^n)={(-1)}^n$, but also we explicitly see that $\det(Q^n) = F_{n+1}F_{n-1} - {F_n}^2$.

So: $F_{n+1}F_{n-1} - {F_n}^2 = {(-1)}^n$.

score 1 · Answer 2 · answered Mar 01 '21 at 09:19

1

This is a known identity:

$F_{n-1}F_{n+1} = F_n^2 + (-1)^n$.

answered Mar 01 '21 at 09:19

Wuestenfux

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multiply two Fibonacci numbers

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