No 4 Fibonacci Numbers can be in Arithmetic Progression
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.
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 $j - 1 = k - 2 = l - 3$.
This theorem requires a proof.
It suffices to show that it must be the case $(a,b,c,d) = (0,1,2,3)$
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Proof Wanted
Arithmetic Progressions
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