# Fibonacci Number is not Product of Two Smaller Fibonacci Numbers

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## Theorem

Let $m, n \in \Z$ be integers.

Let $F_m$ and $F_n$ be the $m$th and $n$th Fibonacci numbers.

Then $F_m \times F_n$ is not a Fibonacci number.

## Proof

We have that:

- $F_n = F_{k - 1} F_{n - k + 2} + F_{k - 2} F_{n - k + 1}$

for $2 \le k \le n$.

Aiming for a contradiction, suppose $F_n = F_m F_k$ for some $m, k \ge 3$.

Then:

\(\displaystyle F_m\) | \(=\) | \(\displaystyle \dfrac {F_n} {F_k}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \dfrac {F_{k - 1} F_{n - k + 2} + F_{k - 2} F_{n - k + 1} } {F_k}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \paren {\dfrac {F_{k - 1} } {F_{k - 1} + F_{k - 2} } } F_{n - k + 2} + \paren {\dfrac {F_{k - 2} } {F_{k - 1} + F_{k - 2} } } F_{n - k + 1}\) |

The right hand side is a weighted mean of $2$ consecutive Fibonacci numbers.

Thus:

- $F_{n - k + 2} < F_m < F_{n - k + 2}$

which cannot happen.

The result follows by Proof by Contradiction.

$\blacksquare$

## Sources

- 1993: Joseph F. Stephany:
*Problems*(*Math. Mag.***Vol. 66**: 61) www.jstor.org/stable/2690479

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