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 sequence.

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 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:

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 $0, 1, 2, 3$.