Prove that that exists a perfect square and perfect cube in the infinite series of Fibonacci numbers?

Question

{1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 91 ...}

How can I go about to prove that there exists or there doesn't exist a perfect square and perfect cube in the series. If there exists these numbers, how can i possibly find the smallest one?

obviously, exclude 1 :)

Answer 1

First, there is a mistake in your question. 91
should be 89.
Next, start hunting! You already have a cube,
F_6 = 8. Also 55+ 89 = 144 = 12².
Here's what's known about these questions:
1,1 and F_12 = 144 are the only squares in the
Fibonacci sequence.
This was proved by J.H.E Cohn in 1964.
1,1 and F_6 = 8 are the only cubes.
This was proved by Yours Truly in 1967.
It was part of my doctoral dissertation.
You can find a simplified proof of this in my paper
On Nth Powers in the Lucas and Fibonacci Series, Fibonacci
Quarterly,16.5 (1978) 451
Very recently, it was proved that
there are no higher powers in this sequence except
1 and 1.
This proof is exceedingly complicated
and relies on estimations for linear forms
in the logarithms of 3 algebraic numbers.
Hope that helps!

Answer 2

For the cubes you can do it if you're willing to:

1) Use the identities:

F2n = Fn(Fn+2F(n-1) )

F(2n+1) ={ F(n+1)}^2 + (Fn}^2

2) Solve in integers: b^6 -20c^6 = 1 This pertains to the first
identity.

3) Use the second identity and the gaussian integers to
solve :

x^3 = F(2n+1) , where x=d^2+e^2

4) This creates for you a second equation in integers to
solve for d and e similar in degree to equation 2).

To do these you have to use algebraic expressions for
consecutive fibonacci numbers and keep on adding them
to get to F(2n+1) , or keep subtracting to get to F(1) = 1.
This accounts for the degree 6 that these diophantine
equations have and they are manageable.

This is a great project and good luck with it.

Hello, Steiner
I think that i can use the gaussian integers to get Fn does
not equal x^m for m>3. Has this approach been done on
this problem since it was first solved? Thank you for your
thoughts.

Here's an example of what i mean: Let Fh be the smallest
fibonacci number which is an mth power for m>3. Suppose that h is even and set h=2n. Then

F2n = Fn ( Fn + 2F(n-1) ) = x^m. If Fn is odd, then the two

factors on the left are coprime and then Fn is an mth power,

contradiction the hypothesis that F2n was the least such.

This means that Fn is even and then the two factors,

2 (Fn) * ((Fn)/2 + F(n-1) ) are coprime which renders them

m th powers again. Set,

2 Fn = a^m and (Fn)/2 + F(n-1) = b^m we get

Fn = 1/2 *a^m

F(n-1)= b^m - 1/4 * a^m and

F(n+1)= b^m + 1/4* a^m

Using the relation for fibonacci numbers:

| F(n-1)F(n+1) - (Fn)^2 | = 1 ,

we get a diophantine equation,

b^(2m) - 5/16 * a^(2m) = 1, or -1

It looks doable. Suppose that now, h is odd in Fh so that

h = 2n+1 and

F(2n+1) = (F(n+1))^2 +( Fn)^2 = x^m. Here, m is odd

since even m renders F(2n+1) a square: 1,144 which are

not mth powers for m>3 (did i remember to exclude 1 ? ).

x is then the sum of two squares : x = r^2 + s^2 and we

may assign real and imaginary parts in the relation,

F(n+1) + i Fn = (r + si)^m.

We get formulas for F(n+1) and Fn in terms of r and s

plus relationships like r divides F(n+1) and s divides Fn

plus another diophantine equation to solve. My apologies

for this brief resurgence of my fibonacci phase : )

Answer 3

I second what the others have said. The problems are not exactly easy and you can look up Cohn's paper as Steiner
mentioned.
I was just doodling with the numbers and here is what i got:

1^3+1^3= 1+1
1^3+2^3= 8+1
2^3+3^3=34+1
3^3+5^3=144+8
5^3+8^3=610+27
8^3+13^3=2584+125

Then i put in the indexes for the fibonacci numbers and the
cubes on the far right to get the general statement:

(Fn)^3 + (F(n+1))^3 = F(3n) + (F(n-1))^3.

That is if you want to prove Steiner's result ! I havent seen the paper. Steiner, could you tell us where to look it up. I
have difficulty finding some of these papers without paying.

Answer 4

1 is a perfect square and a perfect cube and is the smallest one in the series.

Answer 5

Well, the trivial answer is:

1 is the smallest perfect square and perfect cube in
the sequence.