No 4 Fibonacci Numbers can be in Arithmetic Sequence

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.

, 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:

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

So the equality holds.

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