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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?

Let \(m\) and \(n\) be positive integers. What positive integers can be written as \(m+n+\gcd(m,n)+\text{lcm}(m,n)\), for some \(m\) and \(n\)?