Fibonacci Number is not Product of Two Smaller Fibonacci Numbers

Theorem

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

Suppose $\size m, \size n \ge 3$.

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

From Fibonacci Number in terms of Smaller Fibonacci Numbers, 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:

$F_{n - k + 2} < F_m < F_{n - k + 1}$

which cannot happen.

The result follows by Proof by Contradiction.

$\blacksquare$