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.