<pre>
The problem can 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<code>ab; likewise M</code>ab<code>ac</code>bc. Thus
either M and two of a,b,c vanish, or a<code>b</code>c<code>k (say), M</code>k^2, and N=k^3+3k+3.
In the first case, say b<code>c</code>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

<pre>
	     0   k+2   0
	     1    1    k
	     0    1   k+1
</pre>

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

<pre>
	    (k+2)^2=(k+1)(k+3)+1
</pre>

which is only possible if k=0,  Q.E.D.
</pre>
