I need help in proving one elementary result with Fibonacci numbers. Prove that for $n>2$, the product $F_1 \cdot F_2 \cdots F_n$ cannot be a perfect square, where $F_1 = F_2 = 1, F_{n+1}=F_n + F_{n1}$.
1 Answer
Take the largest prime $p$ up to $n$ (which is greater than $n/2$ by Bertrand's postulate) and notice that since the Fibonacci sequence is a divisibility sequence: $gcd(F_p,F_k)=F_{gcd(p,k)}=1$ for $k\neq p$ then if $\prod F_k$ is a perfect square so is $F_p$, but according to this theorem the only nontrivial square in the Fibonacci sequence is 144 and so you get the result.
Remark: The paper "Diophantine equations with products of consecutive terms in Lucas sequences" by F. Luca, T.N. Shorey determines all products of consecutive Fibonacci numbers which are perfect powers.

$\begingroup$ Thank you! Although I am sure that this question should be more elementary than the COHN Theorem, which was open question for decades... $\endgroup$ Commented Sep 13, 2010 at 18:45

1$\begingroup$ Interestingly, from elementery identity F_{n+1}*F_{n1}  F_n^2 = (1)^n it follows (without COHN Theorem) that F_{n+1} and F_{n1} cannot be both exact squares, so if we would have prime p and p+2 greater than n/2, the same agrument would finish the proof ... But twin prime conjecture is a little bit harder than COHN Theorem :) $\endgroup$ Commented Sep 13, 2010 at 18:52

$\begingroup$ I haven't thought about this much, but I couldn't find something simpler. Do you have any reason to believe the problem has an elementary solution? $\endgroup$ Commented Sep 13, 2010 at 19:33

1$\begingroup$ Yes, I have. It is in the book with problems for mathematical olympiads. I prepare highschool students for olympiads and the next topic will be Fibonacci numbers, so I try to find appropriate problems. $\endgroup$ Commented Sep 14, 2010 at 9:37

$\begingroup$ The link to
sciencedirect.com
is broken, but the article can be found at doi:10.1016/j.jnt.2004.08.007 (Zbl 1081.11023). $\endgroup$ Commented May 12, 2023 at 6:06