Prove that there is no bipartite graph on $14$ vertices with degree sequence: $$6, 6, 6, 6, 6, 6, 6, 6, 5, 3, 3, 3, 3, 3.$$
I assume a vertex with degree $5$ breaks this graph from being bipartite.
I was thinking to let the number of vertices with degree $3 = x$ and those with degree $5 = y$. I'm still thinking about what to do from there...
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If I counted correctly, the total degree sum is 68. That means each side of the bipartition must have a degree sum of 34. And 34 is not a multiple of 3 (thus no side can have a degree sum that's a multiple of 3).
Now, take the side of the bipartition that doesn't have the degree 5 vertex. What about its degree sum ?
Because the graph is bipartite, the sum of the degrees of the vertices on the left side is equal to the sum of the degrees of the vertices on the right side. They both are equal to half of the sum of the numbers you are given. You need to show that no subset of those numbers has a sum that is equal to half of the sum of all the numbers. Hint: modular arithmetic.