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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 $F_1 = F_2 = 1, F_{n+1}=F_n + F_{n-1}$.

edited Sep 13 2010 at 18:54

Kevin O'Bryant
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asked Sep 13 2010 at 17:13

Bogdan Grechuk
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Gjergji Zaimi · Answer 1 · 2010-09-13 19:18:39Z UTC

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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 non-trivial 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.

edited Sep 13 2010 at 19:18

answered Sep 13 2010 at 17:31

Gjergji Zaimi
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Thank you! Although I am sure that this question should be more elementary than the COHN Theorem, which was open question for decades... – Bogdan Grechuk Sep 13 2010 at 18:45

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Interestingly, from elementery identity F_{n+1}*F_{n-1} - F_n^2 = (-1)^n it follows (without COHN Theorem) that F_{n+1} and F_{n-1} 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 :) – Bogdan Grechuk Sep 13 2010 at 18:52

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? – Gjergji Zaimi Sep 13 2010 at 19:33

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Yes, I have. It is in the book with problems for mathematical olympiads. I prepare high-school students for olympiads and the next topic will be Fibonacci numbers, so I try to find appropriate problems. – Bogdan Grechuk Sep 14 2010 at 9:37

Elementary problem with Fibonacci numbers

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