Question

F n ≥ 0, let Fₙ denote the n th Fibonacvi number (that is, F₀=0, F₁=1, and Fₙ=F ₙ₋₁+Fₙ₋₂ for all n≥2) What is the greatest possible value of the greatest common divisor of two consecutive Fibonacci numbers?

Answer 1

The greatest possible value of the greatest common divisor (GCD) of two consecutive Fibonacci numbers is always 1.

Step-by-step explanation:

The greatest possible value of the greatest common divisor (GCD) of two consecutive Fibonacci numbers can be found by examining the Fibonacci sequence and identifying patterns. The Fibonacci sequence starts with F₀=0 and F₁=1, and each subsequent term is the sum of the two preceding terms (Fₙ = Fₙ₋₁ + Fₙ₋₂).

When calculating the GCD of two consecutive Fibonacci numbers, we notice that every third term in the Fibonacci sequence is even. Therefore, the GCD of two consecutive Fibonacci numbers is always 1, because consecutive Fibonacci numbers are never both divisible by the same prime number.

For example, consider the Fibonacci sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55...

The consecutive pairs of Fibonacci numbers are: (0, 1), (1, 1), (1, 2), (2, 3), (3, 5), (5, 8), (8, 13), (13, 21), (21, 34), (34, 55)...

The GCD of each consecutive pair is: 1, 1, 1, 1, 1, 1, 1, 1, 1...