Let the penetration of each vertex be ai, and I and I be the number of vertices.

Because it is a complete graph, the in-degree and out-degree of each vertex add up to only (n- 1).

So the degree of each vertex is (n- 1-ai)

If the sum of squares of all vertex degrees is equal to the sum of squares of all vertex degrees

That is, a12+a22+...+an2 = (n-1-a1) 2+(n-1-a2) 2+...+(n-1)

Available by expanding and sorting.

2(n- 1)(a 1+a2+)...+an)=n(n- 1)^2

It's a requirement

a 1+a2+...+an=n(n- 1)/2

In a directed complete graph, the sum of the degrees of vertices does satisfy this condition, so the conclusion holds.

In any directed complete graph, the sum of squares of all vertex degrees is equal to the sum of squares of all vertex degrees.