Let \(n\) be a positive integer. Can \(n^7-77\) ever be a Fibonacci number?
Prove that every pair of consecutive Fibonacci numbers are coprime. That is, they share no common factors other than 1.
Calculate the following: \(F_1^2-F_0F_2\), \(F_2^2-F_1F_3\), \(F_3^2-F_2F_4\), \(F_4^2-F_3F_5\) and \(F_5^2-F_4F_6\). What do you notice?
Work out \(F_3^2-F_0F_6\), \(F_4^2-F_1F_7\), \(F_5^2-F_2F_8\) and \(F_6^2-F_3F_9\). What pattern do you spot?
Can every whole number be written as the sum of two Fibonacci numbers? If yes, then prove it. If not, then give an example of a number that can’t be. The two Fibonacci numbers don’t have to be different.
What’s \(\sum_{i=0}^nF_i^2=F_0^2+F_1^2+F_2^2+...+F_{n-1}^2+F_n^2\) in terms of just \(F_n\) and \(F_{n+1}\)?
What are the ratios \(\frac{F_2}{F_1}\), \(\frac{F_3}{F_2}\), and so on until \(\frac{F_7}{F_6}\)? What do you notice about them?
In the example, we saw that \(\varphi^2=\varphi+1\). Can you write \(\varphi^3\) in the form \(a\varphi+b\), where \(a\) and \(b\) are positive integers?
Among the first \(20\) Fibonacci numbers: \(F_0 = 0,F_1 = 1,F_2 = 1, F_3 = 2, F_4 = 3,..., F_{20} = 6765\) find all the numbers, which have sum of digits equal to their index. For example \(F_1=1\) fits the description, however \(F_{20} = 6765\) does not, since \(6+7+6+5 \neq 20\).