Product of the first $n$ Fibonacci numbers is a perfect square Suppose that $F_{n+2}=F_n+F_{n+1}$ and $F_1=F_2=1$.
Can the number $P_n=F_1\cdots F_n$ be a perfect square if $n\ge 3$?
 A: It is well known that the product $P_n=F_1\cdots F_n$ cannot be a perfect square for $n\ge 3$, see the 
paper "Diophantine equations with products of consecutive terms in Lucas sequences" by F. Luca, T.N. Shorey. It determines all products of consecutive Fibonacci numbers which are perfect powers.
The argument is given also at this MO-question, relying on the fact that $F_n$ is a perfect square for $n\ge 3$ if and only if $n=12$.
A: No,
If $F_n$ is the $n^{th}$ Fibonacci number (assuming we start, $\{1,1,2,3,5\cdots\}$ and not $\{1,2,3,5\cdots\}$) then
$n>2, F_n | F_m \iff n|m$
When p is prime, $F_p$ is co-prime with every Fibonacci number with index $< p$. (? I haven't proven this to myself yet, but it feels right.)
For any $n > 3,$ let $p$ be the greatest prime less than $n.$
$F_p$ is a factor of $P_n$ and $F_p^2$ is not.
A: It is unlikely (and conjectured) that there is a number that is a perfect square and a Fibonacci Number $F_n$ as well, but remains unproven. If this is true, then $P_n$ is not a perfect square.
