We examine the following ECC equation in modular 13: Y2 = X3 + 3X + 8 mod 13 Testing all combinations we harvest eight points: (1,5), (1,8), (2,3), (2,10), (9,6), (9,7), (12,2), (12,11) Together with the added "O" point the ECC set is comprised of nine points. We can construct an 'addition' table showing that adding any two points from the set yields another point from the same set. For example: (2,3) + (9,7) = (1,5) (12,2) + (1,8) = (9,7) (9,6) + (12,11) = (2,10)