If there are n sides, each vertex has (n3) diagonals leading from it (there are n1 other vertices, two of which are its neighbours; a line to any other is a diagonal). So the number of diagonals is n (n3) / 2 (each diagonal can be started from either end). So we have to solven  n (n3) / 2 = 0 n (n3)  2n = 0 n (n  5) = 0 n = 0 or 5, obviously 0 doesn't count. So the answer is 5.
