Alexander Potapenko (of Russia) shows how to go from any example of a polyhedron with n triangular faces to one with n + 2 triangular faces. Connect an arbitrary point on one of its faces to every vertex of this face. We now have 3 triangles instead of 1. Pull out our point. We now have 3 faces instead of 1. So start with a tetrahedron (which has 4 triangular faces) and use the process any number of times to get 6 triangles, then 8, then 10, etc.

Penny Nom (of Regina) sent in a construction that works for any even number of faces except 4: start with a regular n-gon in a horizontal plane, and a point directly above and one below the center of the n-gon. Join each point to the vertices of the horizontal n-gon and you get a polyhedron with 2n triangular faces (n = 3, 4, 5, ...) that is symmetric about the horizontal plane.

These constructions each show that any even number except 2 is possible. Penny next proved that only even numbers are possible. Consider a polyhedron with n triangular faces. Count the number of edges. Each triangle contributes 3 edges to the total, but since each edge belongs to two faces, each edge is counted twice. Thus the number of edges is 3n/2. Since this number is an integer, 2 divides evenly into 3n, which implies that n must be even.

A further comment. Note how our problem is related to the "hand shaking theorem," which says that

At any party, the number of guests who shake hands an odd number of times is an even number.

To see the connection with our problem, imagine a special polyhedral party with one guest standing on each of the triangular faces of the polyhedron, and shaking hands with the guests who are standing on the three neighboring triangular faces. That is, each guest shakes hands 3 times, so the number of guests (which is the number of faces) is even.