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.

$\blacksquare$


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