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Let $G$ be a directed graph with $n$ vertices. Assume that if we forget the directions of the edges of $G$, then we obtain the complete undirected graph $K_n$. Let $v_1, v_2, \ldots, v_n$ be the $n$ vertices of $G$. Prove that: $(od(v_1))^2+(od(v_2))^2+ \cdots + (od(v_n))^2 = (id(v_1))^2+(id(v_2))^2+ \cdots + (id(v_n))^2$ Here, "od" means "outdegree" and "id" means "indegree".

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    This is the 4th graph theory problem (one you deleted) you've asked about in less than 1 day. There's nothing wrong with asking questions, but it gets to the point where it seems like you're just putting up your whole homework. So, if you want help, please tell us what you have attempted first.2012-10-20

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(I prefer to use $\deg_o$ and $\deg_i$ for the out-degree and in-degree.)

HINT: Let the vertices be $v_1,\dots,v_n$; you want to prove that $\sum_{k=1}^n\big(\deg_o(v_k)\big)^2=\sum_{k=1}^n\big(\deg_i(v_k)\big)^2\;,$ or, equivalently, that $\sum_{k=1}^n\left(\big(\deg_o(v_k)\big)^2-\big(\deg_i(v_k)\big)^2\right)=0\;.$

Now $\sum_{k=1}^n\left(\big(\deg_o(v_k)\big)^2-\big(\deg_i(v_k)\big)^2\right)=\sum_{k=1}^n\Big(\deg_o(v_k)-\deg_i(v_k)\Big)\Big(\deg_o(v_k)+\deg_i(v_k)\Big)\;,$ and since the undirected graph is $K_n$, you know exactly what $\deg_o(v_k)+\deg_i(v_k)$ is. You should now be able to reduce the problem to evaluating the sum

$\sum_{k=1}^n\Big(\deg_o(v_k)-\deg_i(v_k)\Big)=\sum_{k=1}^n\deg_o(v_k)-\sum_{k=1}^n\deg_i(v_k)\;,$ which is very easy if you just think about what that last difference of sums really represents.