-1

In the following preprint by Diego Marques, titled "The order of appearance of integers at most one away from Fibonacci numbers", it is claimed that (on page $11$):

It is a simple matter to deduce from Primitive Divisor Theorem that $1, 2$ and $3$ are the only integers which are both Fibonacci and Lucas numbers.

I tried searching for the keyword "Primitive Divisor Theorem" and found out that it was proved (in its simplest form) by Bilu, Hanrot, and Voutier.

Here is my question:

How can I show that $1, 2$ and $3$ are the only numbers which are both Fibonacci and Lucas without using the Primitive Divisor Theorem?

MY ATTEMPT

Using Binet's formulas seem to be out of the question.

Alas, this is where I get stuck!

Bill Dubuque
  • 282,220
Jose Arnaldo Bebita Dris
  • 8,471
  • 1
    So you only tried Binet's formula? How about some simple relations between Lucas numbers and Fibonacci numbers? – Sil Oct 09 '22 at 09:48
  • 1
    @Sil This is in fact the most obvious approach and almost immediately leads to the desired proof. – Peter Oct 09 '22 at 09:55
  • @Peter Your answer seems like the simplest way to handle this (and $L_n=F_{n+1}+F_{n-1}$ is the very first relation on the article about Lucas numbers, that's why I asked about the attempt, Binet's formula is usually a way to make the problem more complicated). By the way I found this https://math.stackexchange.com/questions/28001, guess we could make this a duplicate, but your answer would be nice to keep. – Sil Oct 09 '22 at 10:08

1 Answers1

2

Claim : For every $n\ge 3$ , we have $$F_{n+1}<L_n<F_{n+2}$$ This implies that no Lucas number greater than $3$ can also be a Fibonacci-number. The first inequality immediately follows from $L_n=F_{n+1}+F_{n-1}$

So, it remains to show $F_{n-1}+F_{n+1}<F_{n+2}$ , but this is equivalent to $F_{n-1}<F_{n+2}-F_{n+1}$ hence equivlant to $F_{n-1}<F_n$ which clearly holds for $n\ge 3$

Peter
  • 86,576