To answer the question of proving Fibonacci sequence is periodic mod 5 without using induction., I came across Mathematica to prove $$F_{n}\equiv F_{n+20}\pmod 5$$ for all $n \geq 2$
I defined:
$F[n] := \cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1+\sqrt{5}}{2}\right)^n-\cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1-\sqrt{5}}{2}\right)^n$
and showed:
Simplify[f[n + 20] - f[n] - 10945 f[n] - 6765 f[n - 1]]
equals to 0. (I knew the fact that $F_{n+20} = 10946 F_{n}+ 6765 F_{n-1}$, but I wanted to reach a proof without induction)
Thus, for all $n \geq 2$ we have $$F_{n}\equiv F_{n+20}\pmod 5$$
The problem is that I can not get the intermediate steps of Simplify function, even using the Trace commands. Could you please help?
If possible, I want to also show this in Wolfram Alpha. I don't know how to run several commands in Wolfram Alpha.
Fibonacci[]is built-in. Also,Simplify[]was never meant to show intermediate steps, as what it does under the hood is rather different from what a human would do. – J. M.'s missing motivation Jul 13 '15 at 18:03FunctionExpand[Fibonacci[n]]will give you (a continued version of) Binet's formula. Whatever gave you the idea thatFibonacci[]is unsuitable here? (FWIW, I am somewhat familiar with Binet.) – J. M.'s missing motivation Jul 13 '15 at 18:11