00:01
The graph has order n equal to 3 k plus 3 for some positive integer k.
00:07
Every vertex has degree k plus 1, k plus 2, or k plus 3.
00:11
Prove that g has at least k plus 3 vertices of degree k plus 1, or at least k plus 1 vertices of degree k plus 2, or at least k plus 2 vertices of degree k plus 3.
00:22
So we're going to assume not.
00:34
We'll assume that g has less than k plus 3.
00:42
Vertices of degree k plus one less than k plus one vertices of degree k plus two and less than k plus two vertices of degree k plus three okay so then g has at most most k plus two vertices of degree k plus one uh k plus one not just k vertices of degree k plus two and at most k plus one vertices of degree k plus two okay sense g has three k plus 3 vertices and all have degree k plus 1, k plus 2, or k plus 3.
02:42
If there are a, vertices of degree k plus 1, b, vertices of degree k plus 2, and c vertices of degree k plus 2, and c vertices of degree k plus 3, then a plus b plus c has to be the all the vertices 3 k plus 3 since we just said that a which is the number of vertices of degree k plus 1 is at most k plus 2 b is at most k and c is at most k plus 1 a plus b plus c is at most k plus c is at most k plus 2 plus k plus k plus 1 which is 3 k plus so in particular, we must have that k.
04:10
So a is k plus 2, b is k and c is k plus 1.
04:20
Then the sum of degrees of all vertices is, well, i have k plus 2 vertices of degree k plus 1, i have k vertices of degree k plus 2 and k plus 1 vertices of degree k plus 3.
04:53
So that's equal to k squared plus k and 2k is 3k plus 2 k squared and 2k and 2k and 2k and k squared and k is 4k plus 3k.
05:13
So that's 3k squared plus 3, 4, 5, 9k plus 5.
05:35
If k is even, k squared is even, 3k squared is therefore even, 9k is even...