No 4 Fibonacci Numbers can be in Arithmetic Sequence

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Theorem

Let $a, b, c, d$ be distinct Fibonacci numbers.


Then, except for the trivial case:

$a = 0, b = 1, c = 2, d = 3$

it is not possible that $a, b, c, d$ are in arithmetic sequence.


Proof

Let:

$a = F_i, b = F_j, c = F_k, d = F_l$

where $F_n$ denotes the $n$th Fibonacci number.


Without loss of generality, further suppose that;

$a < b < c < d$

or equivalently:

$i < j < k < l$


Since $i, j, k, l$ are integers, the inequality could be written as:

$i \le j - 1 \le k - 2 \le l - 3$


Now consider:

\(\ds d - c\) \(=\) \(\ds F_l - F_k\)
\(\ds \) \(\ge\) \(\ds F_l - F_{l - 1}\) By assumption, $k - 2 \le l - 3$
\(\ds \) \(=\) \(\ds F_{l - 2}\) Definition of Fibonacci Number
\(\ds \) \(\ge\) \(\ds F_j\) By assumption, $j - 1 \le l - 3$
\(\ds \) \(\ge\) \(\ds F_j - F_i\)
\(\ds \) \(=\) \(\ds b - a\)


For $a, b, c, d$ be in arithmetic sequence:

$d - c = b - a$

This means that the all the inequalities above must be equalities:

$F_l - F_k = F_l - F_{l - 1}$
$F_{l - 2} = F_j$
$F_j = F_j - F_i$


So:

$F_i = 0$

and:

$F_k = F_{l - 1}$
$F_j = F_{l - 2}$

The only Fibonacci numbers having different index but have the same value is $F_1 = F_2 = 1$.

So one of the following is true:

$F_k = F_{l - 1} = 1$
$F_j = F_{l - 2} = 1$
$j - 1 = k - 2 = l - 3$


Suppose the third statement is true.

Write $k = j + 1$, $l = j + 2$.

Then:

\(\ds F_{j + 2} - F_{j + 1}\) \(=\) \(\ds F_{j + 1} - F_j\) $F_j, F_{j + 1}, F_{j + 2}$ form an arithmetic sequence
\(\ds \leadsto \ \ \) \(\ds F_j\) \(=\) \(\ds F_{j - 1}\) Definition of Fibonacci Number
\(\ds \leadsto \ \ \) \(\ds F_j - F_{j - 1}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds F_{j - 2}\) \(=\) \(\ds 0\) Definition of Fibonacci Number

The only zero term of the Fibonacci numbers is $F_0$.

This gives $j = 2$.


Therefore the only arithmetic sequence among Fibonacci numbers satisfying the condition above is:

$F_0, F_2, F_3, F_4$

which corresponds to:

$0, 1, 2, 3$


Now suppose $F_j = 1$.

Since $F_i, F_j, F_k, F_l$ form an arithmetic sequence:

$F_k = F_j + \paren {F_j - F_i} = 2$
$F_l = F_k + \paren {F_j - F_i} = 3$

Which again gives the arithmetic sequence $0, 1, 2, 3$.


Finally suppose $F_k = 1$.

Since $F_i, F_j, F_k$ form an arithmetic sequence:

$F_j = \dfrac 1 2 \paren {F_i + F_k} = \dfrac 1 2$

which is not an integer.

So $F_k \ne 1$.


All cases have been accounted for, and the only arithmetic sequence that can be formed is $0, 1, 2, 3$.

$\blacksquare$


Sources

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