Can 8F2nF2nF6n Be a Square Integer When Fn Is the nth Fibonacci Number?

The Fibonacci sequence, denoted as Fn for n ≥ 1, is defined recursively:

Fn Fn-1 Fn-2 for n ≥ 3 F1 1, F2 1

A more elegant way to describe the n-th Fibonacci number is through a closed-form expression:

Fn (frac{alpha^{n} - beta^{n}}{alpha - beta}) (frac{alpha^{n} - beta^{n}}{sqrt{5}}), where (alpha frac{1 sqrt{5}}{2}) and (beta frac{1 - sqrt{5}}{2}).

Here, (alphabeta 1) and (alphabeta -1). These roots are derived from the polynomial equation (x^2 - x - 1 0).

Decomposition of F2nF2nF6n

Using an identity, we can analyze the product of specific Fibonacci numbers:

Identity: (a^3 - b^3 (a - b)(a^2 ab b^2))

Applying this identity to the Fibonacci sequence:

[sqrt{5} F_{2n} F_{6n} (alpha^{2n} - beta^{2n})(alpha^{6n} - beta^{6n})]

Expanding and simplifying using (alphabeta 1):

[ (alpha^{2n} - beta^{2n})(alpha^{2n} - beta^{2n})^3 cdot 3(alphabeta)^{2n}(alpha^{2n} - beta^{2n})]

[ (alpha^{2n} - beta^{2n})^3 cdot 4(alpha^{2n} - beta^{2n})]

[ (alpha^{2n} - beta^{2n})left((alpha^{2n} - beta^{2n})^2 - 4right)]

[ (alpha^{2n} - beta^{2n})(alpha^{4n} - (beta^{2n})^2)]

Thus:

[5F_{2n} F_{2n} F_{6n} (alpha^{2n} - beta^{2n})^2 (alpha^{2n} beta^{2n})^2]

[ (alpha^{4n} - beta^{4n})^2]

[ (sqrt{5} F_{4n})^2]

Therefore:

Derivation: F2nF2nF6n F4n2

Conclusion

This result tells us that the expression 8F2nF2nF6n will be a perfect square only if the multiplier m is a perfect square itself. Hence, the original claim is proven true:

8F2nF2nF6n 8F4n2 Is a square integer if and only if m is a perfect square.

Key Takeaways

Understanding the closed-form expression of the Fibonacci sequence reveals deeper mathematical relationships. The use of algebraic identities helps in simplifying complex expressions. Proving this result rigorously requires careful manipulation of Fibonacci identities and properties.

Keywords: Fibonacci sequence, perfect square, closed-form expression