Worked solution: Euler's degree argument (1736)
Every bridge has exactly two ends, so building one bridge always adds one to the bridge-count of each of the two landmasses it touches — never to just one of them. Adding up every landmass's bridge-count is therefore exactly like counting handshakes at a party by asking everyone how many hands they shook: every handshake gets counted twice, once from each participant.
Each edge has two endpoints, so it adds exactly to the degree of each of its two vertices — one contribution per endpoint, two per edge. Summing over all vertices therefore counts every edge exactly twice, which is the handshaking lemma: .
For Königsberg's graph this gives , matching from the previous step. Because this sum is always even for any graph whatsoever, it immediately forces the number of odd-degree vertices to be even — — never odd, since an odd number of odd terms would make the total sum odd.
This is a useful sanity check, but by itself it is far too weak to rule out an Eulerian trail: it only tells us that the count of odd-degree vertices is even, and Königsberg's count of is indeed even. The next two steps supply the sharper fact that actually settles the puzzle.