The only perfect powers in the Fibonacci sequence are 0, 1, 8, and 144 

Number theory 1 Minutes

You need to know: Basic arithmetic only.

Background: Let F_0, F_1, \dots, F_n, \dots be the Fibonacci sequence defined by F_0 = 0, F_1 = 1, and F_{n+2} = F_{n+1} + F_n for n\geq 0. Let L_0, L_1, \dots, L_n, \dots be the Lucas sequence defined by L_0 = 2, L_1 = 1, and L_{n+2} = L_{n+1} + L_n for n\geq 0.  A perfect power is an integer of the form m^p for integers m and p\geq 2.

The Theorem: On 24th November 2003, Yann Bugeaud, Maurice Mignotte, and Samir Siksek submitted to the Annals of Mathematics a paper in which they proved that (a) the only perfect powers in the Fibonacci sequence are F_0 = 0, F_1 = 1, F_2 = 1, F_6 = 8, and F_{12} = 144, and (b) the only perfect powers in the Lucas sequence are L_1=1 and L_3=4.

Short context: The Fibonacci sequence is perhaps the most famous and well-studied sequence of integers in mathematics. In 1951, Ljunggren proved that the only perfect squares in this sequence are 0=0^2, 1=1^2, and 144=12^2. In 1969, London and Finkelstein proved that the only perfect cubes are 0,1 and 8=2^3. Are there any other perfect powers in the sequence? By 2003, it was known that there are no more p-th powers with p\leq 17 or with p\geq 5.1\cdot 10^{17}. The Theorem proves this for all p, and resolves this question also for the Lucas sequence.

Links: Free arxiv version of the original paper is here, journal version is here. See also Section 6.3 of this book for an accessible description of the Theorem.

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Published by Bogdan Grechuk

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