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A digraph on $n$ vertices is called a tournament if there is a exactly one directed edge between any two distinct vertices. A vertex $v$ dominates a vertex $w$ if there is an edge from $v$ to $w$.

Let $dm(v)$ be the set of vertices that are dominated by $v$, and $dm(v,w)$ be the set of vertices that are dominated by both $v$ and $w$. Then the digraph is called a regular tournament if there is a positive integer $m$ such that $dm(v)=m$ for all $v$, and strongly regular if there are positive integers $m_1$ and $m_2$ such that $dm(v,w)=m_1$ for all $v \neq w$ and $dm(v)=m_2$ for all $v$.

I am asked to show that if a tournament is regular that $n$ is odd and that if it is strongly regular that $4 \mid n-3$. I am allowed to use the fact that in a regular digraph, for every vertex $v$ the in-degree of $v$ is equal to the out-degree of $v$.

The first part is not too hard: if $G$ is a tournament and we remove a vertex $v$ from $G$ then the in-degree of $v$ is equal to the out-degree so $G/v$ must have an even number of vertices.

I have not been able to solve the second part of the question.

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  • $\begingroup$ "the fact that in a regular digraph, for every vertex $v$ the in-degree of $v$ is equal to the out-degree of $v$" Fact??? This is false for a $4$-vertex digraph that consists of a $3$-cycle and a further vertex that has an edge into the cycle. It is true for tournaments, but for a rather nontrivial reason (indeed, in any digraph, the sum of all indegrees equals the sum of all outdegrees; but in a tournament, it is also true that the sum of the squares of all indegrees equals the sum of the squares of all outdegrees, and thus we are in the equality case of the Cauchy-Schwarz inequality). $\endgroup$ Jan 22, 2019 at 14:51

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Late answer, but it might be useful to someone.

The sum of the out-degrees of all vertices is $n(n-1)/2$ (since a tournament on $n$ vertices has exactly $n(n-1)/2$ edges), so $m_2 = (n-1)/2$.

Now let $A$ be the set of couples of distinct edges directed to a same vertex.

$A = \{(u,w),(v,w) \in E^2, u \neq v\}$

For one vertex $u$, we can consider the set of the couples of edges that come from two different $v$ and $w$ to $u$. As the in-degree of $v$ is $m_2$ (since, as you said, for each vertex $v$, the in-degree is equal to the out-degree), this set has $m_2(m_2 - 1)/2$ elements.

Taking all those sets for every vertex in our graph gives a partition of $A$, thus $A$ has $n\cdot m_2(m_2 - 1)/2$ elements.

An other partition of $A$ is given by the sets $dm(v,w)$ you defined for every couple of vertices. So the number of elements of $A$ is equal to $n(n-1)/2 \cdot m_1$.

We conclude that $n\cdot m_2(m_2 - 1)/2 = n(n-1)/2 \cdot m_1$ which gives $n = 4m_1 + 3$.

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