No 4 Fibonacci Numbers can be in Arithmetic Sequence

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

Then it is not possible that $a, b, c, d$ are in arithmetic progression, except the trivial case:


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

Proof
Let $a=F_i$, $b=F_j$, $c=F_k$, $d=F_l$ where $F_n$ is 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:


 * $i \leq j-1 \leq k-2 \leq l-3$

Now consider:

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

So the equality holds.

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