The proof I saw uses the pigeonhole principle. They list the first $101$ terms beginning with $1,1,2,3,5,...$ and then group them in pairs, $(1,1),(1,2),(2,3),...$, and this gives us $100$ pairs. Now, since the pair $(0,0)$ can't occur, we must have a pair that repeats, and a pair occurring again, will lead to the same sequence.
My question Why is it that in the $100$ pairs, there are only $100$ possible choices of pairs? and getting rid of $(0,0)$ implies that there are only $99$ left to choose from?
Thanks in advance
If we are taking the Fibonacci sequence modulo $10$, then the only possible values the sequence can take are $0$ through $9$. You cannot pick a number that lies outside of this range, so there are only $100$ choices for pairs. How would you possibly pick something that is not a part of these $100$ pairs?
You are correct in that getting rid of $(0,0)$ implies there are only $99$ pairs left to choose from. This doesn't change the logic of the proof, it only makes sure that we don't end up with a degenerate case.