Why is there only one perfect square (144) and one perfect square root (8) in the infinite Fibonacci sequence?

Question

Since the number sequence is infinite, there's a very low chance that there is only one perfect cube and one perfect square in the entire known sequence. Is there a mathematical reason for this?

Answer 1

You are forgetting that there are 3 squares and 3
cubes in the Fibonacci sequence.
F_1 = F_2 = 1 and F_12 = 144.
F_1 = F_2 = 1 and F_6 = 8.
Both these theorems have been proved mathematically.
The proofs are too long to post here, but I can
say that the first one was proved by J.H.E. Cohn
in 1964 and involves the study of quadratic
residues of the numbers in the Fibonacci
sequence.
I proved the second one in my doctoral dissertation.
It involves a study of the cubic field Q(5^1/3).
Incidentally, this theorem also is equivalent
to the class number 1 problem for imaginary
quadratic fields. (A famous problem of Gauss)
Look in the Fibonacci Quarterly from 1964
and 1969 for complete proofs.
Or e-mail me if you would like reprints.

Answer 2

Here is a proof. If you understand it you are a mathematical genius.
http://math.asu.edu/~checkman/SquareFibonacci.html

Answer 3

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