Prove that for n greater than or equal to 2 the n-th Lucas number is equal to $[a^n+1/2]$.
The brackets are the greatest integer function, $a = \frac{1+\sqrt5}{2}$. I get every kind of proof we have done in number theory except the greatest integer function ones, so I am not sure how to even start this one.
Oh, if $a = \phi$, then the question makes sense so I will assume that. Now note that from the recurrence of the Lucas numbers, we can easily find that $L_n = \phi^n + (1-\phi)^n$ from the roots of the characteristic quadratic.
Then note that all Lucas numbers are integers. Furthermore, $(1- \phi)^n < 1$. Therefore, $L_n$ is the nearest integer to $\phi^n$. So the problem reduces to proving that the nearest integer to a real number $x$ is $\lfloor(x+1/2 \rfloor$ which just falls from casework.