We will proceed by contradiction and induction.
Assume that there is no loop. If two bridges connect the same pair of islands, then we have a loop. So, we cannot have two bridges connecting the same islands.
Case 1: Assume now that each island has two bridges. Denote the islands as . WLOG, connects to , connects to , and so on. However, this yields a contradiction since and both have one bridge. To fix this, we can connect them together, which results in the whole graph being a cycle of length . On the other hand, we now have islands and bridges. Uh-oh. Let's try to fix this by adding in an arbitrary bridge connecting and . Now note that we have a cycle connecting which is a contradiction. Thus, if each island has at least two bridges, then there must be (at least) a cycle.
Case 2: If there is an island with less than \(2\) bridges, then we can "ignore" the island and all bridges extending out of the island. We can keep performing this operation until all islands have at least \(2\) bridges. Now, by Case 1, we are done.
Otherwise; answered at https://artofproblemsolving.com/community/c243175h1500558_combinatorics_2_induction
Max that sounds like total nonsense.
If your teacher did not want other kids to cheat they would not want you to cheat either!
You take if off to attempt to disguise the fact that you are cheating.
It is extremely rude. People have been banned from this site for doing this.
People go to the effort of answering your question and then you blank your question. You have not even bothered to give Olpers a proper thankyou.
You have not even given him/her a point although I notice you are quite happy to give yourself a point just for asking a question.
It is not good that you have drawn attention to yourself in this way.
Do not ever delete any of your question in the future!