I just came to know about Pisano periods, and it's amazing application in computing modulus of large Fibonacci numbers. I know that a Pisano period periodically starts at $0,1$, but I haven't been able to figure out why this pattern occurs periodically.
Is there any underlying concept behind this pattern, or was it just pure luck that it was discovered?
On
Although this is not perhaps the mathematical reason you were looking for, this table does provide an idea of how Pisano periods come about. You can see that the Fibonacci numbers $F_i$ have been computed and also the $F_i\,\mathrm{mod}(m)$. You can see that the numbers for a particular $F_i\,\mathrm{mod}(m)$ is periodic. So for example, for $F_i\,\mathrm{mod}(2)$ the numbers 011 repeat. Similarly, 01120221 repeats for $F_i\,\mathrm{mod}(3)$. There is a conjecture that for $n > 2$, the Pisano period is even. That is, it contains an even number of digits. You can look into that as well
And also all Pisano numbers start from 01 as you can see
We consider the Fibonacci sequence with $f_0=0,f_1=1$ and $f_{n+2}=f_{n+1}+f_n\;\forall\;n\geq0$. Consider the pairs of consecutive Fibonacci numbers $(f_k,f_{k+1})$ modulo $n$. There are only $n^2$ possible pairs like this. Hence by pigeon hole principle $\exists$ two distinct pairs $(f_r,f_{r+1})$ and $(f_s,f_{s+1})$ which are identical modulo $n$. Let $r>s$. Then $$f_r\equiv f_s\pmod{n}$$ $$f_{r+1}\equiv f_{s+1}\pmod{n}$$ which implies $$f_{r+2}=f_r+f_{r+1}\equiv f_s+f_{s+1}=f_{s+2}\pmod{n}$$ and $$f_{r-1}=f_{r+1}-f_r\equiv f_{s+1}-f_s=f_{s-1}\pmod{n}$$ proceeding this way we get that $$f_{r-s}\equiv f_0\pmod{n}, f_{r-s+1}\equiv f_1\pmod{n}$$ hence the sequence $f_n$ is periodic with period $r-s$.
Moreover let $$\mathcal{F}=\begin{bmatrix} 1&1\\1&0 \end{bmatrix}$$ in $GL_2(\mathbb{Z/n\mathbb{Z}})$
If $\pi(n)$ is the Pisano period modulo $n$ then $$\mathcal{F}^{\pi(n)}=I$$
From this we can conclude that the Pisano period is even. $$\mathrm{det}(\mathcal{F}^{\pi(n)})=(-1)^{\pi(n)}=1$$ which implies $\pi(n)$ is even.