Translation to a graph theoretic problem

Suppose , $G(V,E)$ be a simple graph with $\boxed{k=100}$ vertex. Suppose $V=\{v_1,\cdots ,v_k\}$ And we connect $v_i,v_j$ if they are friend $i\neq j$ . Suppose $a_i = \deg (v_i)$ .And $\tau _j$ denote number of vertices having degree more than $j,0\le j \le k-1$ .

We would like to show that $\sum_{i=1}^k a_i = \sum_{j=0}^{k-1} \tau_j$ .

$\boxed{\text{Key Lemma}}$ .
$\sum_{j=0}^{k-1} \tau _j =2|E|$ .
Proof.
Suppose the statement is true for $k=n$ .
Then join a certex call $v_{n+1}$ .If it is isolated then $\sum_{i=0}^n \tau ' _i = \sum _{i=0}^{n-1} \tau_i$ , $\tau '_j$ count number of vertex having degree more than $j$ in the new graph with $n+1$ vertex. we are done.

if $v_{n+1}$ is not isolated. . Suppose it joins with $v_j$ , $j\le n$ . Then $\deg (v_{n+1})=1$ .
And , $\tau ' _{\deg v_j} = \tau_{\deg v_j} +1$ ,
$\tau '_0 =\tau_0 +1$ .
And another $\tau_j$ will as same as $\tau '_j$ .

So,in this case $\sum_{j=0}^{n} \tau '_j = 2|E|+2$ .One must see that number of edges in new graph is $|E|+1$ .

Similar approach can prove the case for more than $1$ new edges $\blacksquare$ .

Now , $\sum a_j = \sum_{v\in V} \deg (v)=2|E|$ .

So, $\sum_{i=1}^k a_i = \sum_{j=0}^{k-1} \tau_j$ .

Proved!!!!