# 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 $j - 1 = k - 2 = l - 3$.

## Sources

- 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $5$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $5$