==> induction/party.s <==
Here is an easy solution by induction. Let P be the set of people in the
party, and n the size of P. If n=2 or 3, the result is trivial. Suppose now
that n>3 and that the result is true for n-1.

For any two distinct x,y in P, write x & y to mean that `x is a friend of y',
and x ~& y to mean that `x is not a friend of y'.

Take q in P. The hypothesis on the relation & is still satisfied on P-{q}; by
induction, the result is thus true for P-{q}, and there is some p in P-{q}
such that p & x for any x in P-{p,q}. We have two cases:

a) p & q. Then the result holds for P with p.

b) p ~& q. By hypothesis, there is a unique r in P-{p,q} such that p & r & q.
For any x in P-{p,q}, if x & q, then p & x & q, and so x=r. Thus r is the
unique friend of q. Now for any s in P-{q,r} there exists some x such that s &
x & q, and so x=r. This means that r & s for any s in P-{q,r}, and as r & q,
the results holds in P with r.

The problem can also be solved by applying the spectral theory of graphs
(see for instance Bollobas' excellent book, _Extremal Graph Theory_).

The problem's condition is vacuous if there is only N=1 person at the "party", 
impossible if N=2 (If you aren't your own friend, nor I mine, somebody *else*
must be our mutual friend), and trivial if N=3 (everybody must be everyone
else's friend).  Henceforth assume N>3.

Let A,B be two friends, and C their mutual friend.  Let a be the number
of A's friends other than B and C, and likewise b,c.  Each of A's friends
is also friendly with exactly one other of A's friends, and with none of
B and C's other friends (if A1,B1 are friends of A,B resp. and of each other
then A1 and B have more than one mutual friend); likewise for B and C.
Let M=N-(a+b+c+3) be the number of people not friendly with any of A,B,C.
Each of them is friendly with exactly one of A's and one of B's friends;
and each pair of a friend of A and a friend of B must have exactly
one of them as a mutual friend.  Thus M=ab; likewise M=ab=ac=bc. Thus
either M and two of a,b,c vanish, or a=b=c=k (say), M=k^2, and N=k^3+3k+3.
In the first case, say b=c=0; necessarily a is even, and A is a friend of
everybody else at the party, each of whom is friendly with exactly one other
person; clearly any such configuration (a graph of k/2+1 triangles with a
common vertex) satisfies the problem's conditions).

It remains to show that the second case is impossible.  Since N=k^2+3k+3
does not depend on A,B,C, neither does k, and it quickly follows that the
party's friendship graph is regular with reduced matrix

	     [  0   k+2   0  ]
	     [  1    1    k  ]
	     [  0    1   k+1 ]

and eigenvalues k+2 and +-sqrt(k+1) and multiplicities 1,m1,m2 for some
*integers* m1 and m2 such that (m1-m2)*sqrt(k+1)=-(k+2) (because the graph's 
matrix has trace zero).  Thus sqrt(k+1) divides k+2 and k+1 divides

	    (k+2)^2=(k+1)(k+3)+1

which is only possible if k=0,  Q.E.D.