It really depends on how you define things. As Elaqqad said, you can't get rid of induction when you define the sequence by Induction. But if you define it from the start with no induction, you can prove some of its features without induction. So, be pragmatic. But, as you wish, I do prove **Fibonacci sequence is periodic mod 5 without using induction**.

Let's start from a non-inductive definition of Fibonacci sequence. I use Binet formula, as a closed-form expression of the Fibonnaci number.

Fibonnaci number is defined as a function of $n$, which outputs an _Integer_, and defined as*:

$F[n] := \cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1+\sqrt{5}}{2}\right)^n-\cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1-\sqrt{5}}{2}\right)^n$

We are about to show $$F_{n}\equiv F_{n+20}\pmod 5$$ for all $n \geq 2$

Actually, we also calculate the reminder. We prove that:

$F[n+20]-F[n]=10945.F[n]+6765F[n−1]$

Doing this is really cumbersome. You can't do this by hand, but using a computational software program like Mathematica, you can do it easily:

Do this in Mathematica:

    Simplify[f[n + 20] - f[n] - 10945 f[n] - 6765 f[n - 1]]

and you get:

    0

Done. :-)

I'm trying to extract the exact intermediatory simplifying steps from Mathematica. I'll post it as soon as I get it.

* I am not going to prove that F(n) is actually an integer here.