Olpers

I don't know how any math could be fun, but I'll take your word for it. Lol

Nevermind, I understand now

I'm a bit confused, would I expand the two polynomials then multiply the answer all by 10?

Thanks

Thanks both of you!

Thanks!

Haha I didn't really answer the question, but yeah I agree with you.

Haha why did you delete the question?

Thank you! The picture is very clear!

If you assume such a constant exists, then you can solve for it by setting up an equation based on P(x, y, z) = 0 whenever x + y + z = 0. For example, P(2, -1, -1) = 0.

Thank you!!! 

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.