No 4 Fibonacci Numbers can be in Arithmetic Progression

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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 progression.


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:

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


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

$d - c = b - a$

So the equality holds.


So $a = 0$ and $j - 1 = k - 2 = l - 3$.



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