I know you, you know me

I am not sure whether I made any mistakes in my approach. Let me know if this makes sense.

Consider the first conjecture.

1) The statement simply says that for any group of size 2n or 2n+1,

• if for all subgroups of size n, at least 1 person outside the subgroup knows all the people in the subgroup, then there is at least one person who knows all others in the group.

Assume that there is no person who knows all others in the group i.e, for all the people in the group, there is at least 1 person in the group, that they don't know (Assuming the group is not 1-acquainted). So there will be at least n pairs of people who don't know each other ( Let us assume the group size is 2n for now). Let us arrange them in the following order

• $1^{st}$ and $(n+1)^{th}$ person doesn't know

• $2^{nd}$ and $(n+2)^{th}$ person doesn't know

• .

• .

• $n^{th}$ and $2n^{th}$ person doesn't know.

Now consider the subgroup from $1^{st}$ to $n^{th}$ person. Clearly there won't be a person outside the subgroup who knows all the people in the subgroup. For every $k^{th}$ person outside the subgroup, n<k<=2n, there will be a $(k-n)^{th}$ person, who is present in the subgroup and doesn't know the $k^{th}$ person. Hence the group is not (n,1)-amicable. For group size of 2n+1, we can simply assume that $n^{th}$ , $2n^{th}$ and $(2n+1)^{th}$ person doesn't know each other and proceed with the other steps as it is. Hence it is proved that if the group is not 1-acquainted, it is not (n,1)-amicable.

Taking the contrapositive, $\boxed{(n,1)-amicable => 1-acquainted.}$

Now consider the second conjecture.

2) Assume that the group is not 2p-acquainted, i.e, there is no subgroup of 2p people such that everyone outside the subgroup knows at least one of them. Now consider an arbitrary subgroup of size 2p. Let us name it S. Consider the group as a single person. The person S knows another person iff at least one of the elements of S knows the person and vice versa. Thus the group size is now 2n-2p+1 = 2(n-p)+1 . Since the group is not 2p-acquainted, it is neither 1-acquainted. Comparing this to the first conjecture, we can deduce that the group is not (n-p,1)-amicable, i.e, there exists a subgroup of size (n-p) such that it is not known by any person. From the proof of conjecture 1, it is clear that there exits a subgroup in such a way that person S is not a part of the subgroup. Note that any group which is not (n,1)-amicable is also not (n,p)-amicable where p>=1. Thus the group is (n-p,2p)-amicable. Hence not 2p-acquainted implies not (n-p,2p)-amicable.

Taking the contrapositive, $\boxed{(n-p,2p)-amicable => 2p-acquainted.}$

Guesswork. Does wonders for you. Both correct