Try to give a complete proof.

1. Quoted from MTN

设这群人总数为N。因为N有限，所以总能找出一个朋友最多的人，设为a0。a0共

有k个朋友，设为a1,a2,...,ak。显然，a1,a2,...,ak中任意两人的共同朋友是

a0。而因为a0仅有这k个朋友，所以a0和a1,a2,...,ak中任意一人的共同朋友也只

能在这k个人中。因此a1,a2,...,ak中任意一人在这k个人中有且只能有一个朋友，

不然如果a1有两个朋友a2和a3，那么a2和a3就有两个共同朋友a1和a0。这就推出

一个结论，k一定是偶数。

From now on, let A = (a0, a1, ..., ak).

The structure of A=(a0, a1, ..., ak) is a FAN, mentioned in Professor W's post.

2. quoted from MTN.

现在假设a0并没有认识所有人，即k<N-1，那么，至少有一个人不在这k+1个人中，

设这个人为b0。b0和a0的共同朋友只能在a1,a2,...,ak中，假设是a1。除此之外，

b0在a1,a2,...,ak中不能再有第二个朋友，因为如果还有个朋友aj的话，a1和aj

就有a0和b0两个共同朋友。由此推出，b0与a1,a2,...,ak中任意一人的共同朋友

只能在a1,a2,...,ak之外。而且，由于a1,a2,...,ak中任意两人已经有共同朋友

a0，因此a1,a2,...,ak中任意两人在a1,a2,...,ak之外不能有共同朋友。

Let f(ai) is the commone friend of b0 and ai. Then, f(ai) is outside A, and ai != aj implies f(ai) != f(aj). Indeed, if bk = f(ai) = f(aj) and ai != aj, then bk and a0 have two friends: ai, and aj, this is a contradiction. Let B=(b0, a1, b2, ..., bk), where bj = f(aj).

All a1, b2, ..., bk are friends of b0. and b2, ..., bk are outside A.

Since inside B, b0 already have k friends, b0 should not have more friend. A conclusion is that any member outside A should have only one friend in A.

3. Here is the contradiction.

b0 and a1 must have a friend in B, say who is b2. Then from the definition b2 = f(a2). Now b2 is outside A, but have two friends a1 and a2 inside A.