Regular graph
Task number: 4173
Show that if the \( 2k \)-regular graph has an even number of edges, then either \( k \) or \( | V_G | \) is even.
Solution
If we count the degrees, we get twice the total number of edges, that is \( | E_G | = \frac{2k | V_G |}2 = k | V_G | \).
If both factors were odd, we would get an odd number of edges, which would contradict the assumptions.
