Now this question has been asked in this site and I am aware of that. But after some reading I still don't get the answer there.
This is a field with finitely many elements. I am asked to prove that a natural number $n$ is present such that:
based on other answers from quora: https://www.quora.com/Suppose-F-is-a-field-with-finitely-many-elements-How-do-I-prove-that-there-exists-a-natural-number-n-such-that-1-1-1-1-1-n-times-0
Almost everyone applies that pigeon hole principle. Many of the answers included said something like: "there exists two elements in the sequence which equal each other."
I really don't see where (at least) two elements are equal to each other in the sequence? I am really confused on this part. I understand the lines after that (for example you can use the other element as an additive identity, if thats so, it is a field because $0$ - the unique element is present. But I don't get the two elements are equal part? How can I see this? Can someone provide another example in terms of math? Because I have read the wikipedia definition and that only make sense intuitively.
So, if I understand the question correctly, you have a fiinite field $F$ and you want to prove that there exists an $n\neq0$ such that $$ n\cdot1_F=\underbrace{1_F+\cdots+1_F}_n=0_F $$ using the pigeonhole principle.
I would do as follows:since $|F|<\infty$ for $k>0$ sufficiently large you have to find some repetitions in the sequence $$ 1_F,\quad 2\cdot1_F,\quad 3\cdot1_F,\quad...,k\cdot1_F. $$ This is because the "pigeons", i.e. the numbers $\{1,...,k\}$, are more than the "holes", i.e. the elements of $F$.
So, let $r<s$ such that $r\cdot1_F=s\cdot1_F$. Then if $n=s-r>0$ we have $n\cdot1_F=0$.
I must say that this is a unusual way to prove the claim. The usual strategy is to observe that there's a canonical ring homomorphism $$ \phi:\mathbb{Z}\rightarrow F,\qquad\phi(k)=k\cdot1_F $$ and the finiteness of $F$ implies that $\ker(\phi)$ is a non trivial ideal in $\mathbb{Z}$, so that $\ker(\phi)=n\mathbb{Z}$ for some $n\geq0$ that in fact can be assumed positive and shown to be prime.