Difference between revisions of "Divisibility of Fibonacci numbers"
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Using the fact that <math> A^{m+n} = A^m \times A^n </math> and comparing the (1,2)-entries of the two matrices, we obtain a helpful representation of <math> F_{m+n} </math>: | Using the fact that <math> A^{m+n} = A^m \times A^n </math> and comparing the (1,2)-entries of the two matrices, we obtain a helpful representation of <math> F_{m+n} </math>: | ||
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| + | <math> F_{m+n} = F_{m-1} F_n + F_{m} F_{n+1}. </math> | ||
Revision as of 21:21, 20 December 2021
Recall that Fibonacci numbers are the members of the sequence 1, 1, 2, 3, 5, 8, ..., so that every next number is the sum of two previous numbers. It is a part of the mathematical folklore that every 5th Fibonacci number is divisible by 5. It also happens that 5 is the 5th Fibonacci number. In fact, a more general statement is true:
Every n-th Fibonacci number (i.e. number [math]\displaystyle{ F_{nk} }[/math] for some k) is divisible by [math]\displaystyle{ F_n }[/math], the n-th Fibonacci number.
Among the different ways to prove that is to use the matrix
[math]\displaystyle{ A = \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} }[/math]
It is easy to verify by induction on [math]\displaystyle{ n }[/math] that
[math]\displaystyle{ A^n = \begin{bmatrix} F_{n-1} & F_n \\ F_n & F_{n+1} \end{bmatrix} }[/math]
Using the fact that [math]\displaystyle{ A^{m+n} = A^m \times A^n }[/math] and comparing the (1,2)-entries of the two matrices, we obtain a helpful representation of [math]\displaystyle{ F_{m+n} }[/math]:
[math]\displaystyle{ F_{m+n} = F_{m-1} F_n + F_{m} F_{n+1}. }[/math]