Graph theory problem - the number of common acquaintances This problem is from the book Graph Theory by Bin Xiong. Although I tried to understand the authors answer, their explanation is unclear for me. The problem is as follows.

There are n people $A_{1}, A_{2}, ...,  A_{n},$  taking part in a mathematics contest, where some people know each other and any two people  who do  not  know  each  other  would  have  common acquaintance. Suppose that $A_1$, and $A_2$ know each other, but do not have common acquaintance.  Prove that the acquaintances of $A_1$ are as many as those of $A_2$.

The authors are proving the problem by denoting $n$ people by $n$ vertices and putting an edge between any two vertices if the corresponding two people know each other.
My questions are:


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*I am wondering if there are some conditions which the problem statements lack. For example, if there are 3 people  $A_{1}, A_{2}, A_{3}$ and suppose that $A_{1}, A_{3}$ know each other,  the number of acquaintances of $A_1$ and $A_2$ does not seem to be same.

*And the authors are pointing out that there are no any 2 nonadjacent vertices which have 3 common neighbors. How could we infer this condition? Does the problem statements specify this condition explicitly?
 A: If the condition "any two people who do not know each other would have common acquaintance" is strengthened to read "any two people who do not know each other have exactly two common acquaintances", then the statement is correct and has the following easy proof.
For $i=1,2$ let $\mathcal N_i$ be the set of all acquaintances of $A_i$ in $\{A_3,\dots,A_n\}$. By hypothesis $\mathcal N_1$ and $\mathcal N_2$ are disjoint. If $X\in\mathcal N_1$ then $X$ and $A_2$, who are not acquainted, have exactly two common acquaintances; one of then is $A_1$, the other one is in $\mathcal N_2$. Thus each member of $\mathcal N_1$ has exactly one acquaintance in $\mathcal N_2$, and vice versa. Hence $\mathcal N_1$ and $\mathcal N_2$ have the same number of elements, call it $k$, and so each of $A_1,A_2$ has exactly $k+1$ acquaintances.
More generally, if there is an integer $m\gt1$ such that any two nonadjacent vertices have exactly $m$ common neighbors, then any two adjacent vertices with no common neighbors have the same number of neighbors.
