Prove that every 4th Fibonacci number is divisible by 3, that is 3|f4n, for all n>=1.
Induction and some patience:$\displaystyle F_{4n}=F_{4n-1}+F_{4n-2}=F_{4n-2}+F_{4n-3}+F_{4n-3}+F_{4n-4}$ $\displaystyle = F_{4n-3}+F_{4n-4}+F_{4n-3}+F_{4n-3}+F_{4n-4}=...$
Tonio
Pd. The solution's already present in the rightmost expression above.