An n-Venn diagram is a Venn diagram on n sets, which is defined to be a collection of n simple closed curves (Jordan curves) C1,C2, . . . ,Cn in the plane such that any two intersect in finitely many points and each of the 2n sets of the form ∩C i i is nonempty and connected, where i is one of “interior” or “exterior.” Thus the Venn regions are all bounded except for the region exterior to all curves; each bounded region is the interior of a Jordan curve. See [6] for much more information on Venn diagrams. An n-Venn diagram is symmetric if each curve Ci is ρ i (C1), where ρ is a rotation of order n about some center (we use O for the fixed point of rotation ρ). We use Boolean notation for combinations of sets, with the 0-1 string e1e2 . . . en representing ∩C i i , where i is interior (respectively, exterior) if ei = 1 (respectively, 0). Thus 111 . . . 1 represents F , the full intersection of all the interiors, 000 . . . 0 is the intersection of all the exteriors (the unbounded region), and 100 . . . 0 represents the set of points interior to C1 and exterior to the others. In a symmetric Venn diagram, rotation of a region by ρ corresponds to a rightward cyclic shift of the Boolean string. The universally familiar three-circle Venn diagram is symmetric, as is the one on two sets using two circles. For about 40 years a major open question was whether symmetric n-Venn diagrams exist for all prime n. Henderson found one for n = 5 and also (unpublished) for n = 7. Much later, Hamburger [3] settled the case of 11, which was quite complicated, and then in 2004 Griggs, Killian, and Savage [1] found an approach that works for all primes. So we now have the strikingly beautiful theorem that a symmetric n-Venn diagram exists if and only if n is prime. But there is a small problem: Henderson’s proof, which appears to be very simple, has a gap. Here is the proof from [4]. Suppose 1 ≤ k ≤ n − 1. Since a symmetric n-Venn diagram is symmetric with respect to a rotation of 2π/n, the regions corresponding to the Boolean strings with k 1s must come in groups of size n, each group consisting of one such region and its images under repeated rotation by 2π/n. Therefore n divides (n k ) . This concludes the proof because the only n for which this is true for the specified k-values are the primes (an easy-to-prove fact of number theory; see [5]). This is a very seductive argument. The primeness arises in such a cute way that one wants it to be true. Thus the proof has been repeated in many papers in the decades since it was first published. Yet there are problems. The proof does not call upon the connectedness of the Venn regions. Without connectedness the result is false; see Figure 1 (due to Grunbaum [2]), which shows a diagram satisfying all of the conditions