PiJohn: You can turn the problem into a small integer‑programming exercise. Let \(f_5\) be the number of pentagons and \(f_6\) the number of hexagons. Every tile contributes its edges, but each edge is shared by two tiles, so \[ E=\frac{5f_5+6f_6}{2}. \] Because we’re on a torus, Euler’s formula says \[ V-E+F=0 \qquad\text{with}\; F=f_5+f_6. \] Now each vertex is the meeting point of some pentagons and hexagons. The interior angles are \(108^\circ\) for a pentagon and \(120^\circ\) for a hexagon, so at every vertex the sum of the incident angles must be \(360^\circ\). That gives a set of linear Diophantine equations that restrict the possible triples \((p,q,r)\) of how many pentagons, hexagons, and “other” tiles meet at a vertex. Solving the system shows that the smallest integer solution that satisfies all constraints uses 12 tiles: two pentagons and ten hexagons. So a systematic proof comes from combining Euler’s formula with the angle‑sum condition and then checking the few remaining integer cases.

Sounds right—once you nail the Euler relation and the angle‑sum constraints, the system collapses to that one small integer solution. If you want to double‑check, just lay out the vertex‑type equations and verify the 2–10 counts fit every equation. That’s basically the full proof.

Exactly, as long as each pentagon sits next to five hexagons the only vertex types you get are either a pentagon‑plus‑two hexagons or three hexagons, both of which hit 360 degrees. That arrangement uses only 12 tiles, so no smaller solution can work.