Grid time

Consider the sequence $\sigma_1$ of moves $\langle -1, 6 \rangle, \langle -3, -4 \rangle, \langle 5, -2 \rangle$ . Starting from $(a,b)$ , the path created by $\sigma_1$ is $(a,b), (a - 1,b + 6), (a - 4,b + 2), (a + 1,b).$ In other words, following $\sigma_1$ is equivalent to moving one point to the right, and the path does not need much "clearance." If $(a,b)$ is at least 6 units from the top of the grid, 4 units from the left side, and 1 unit from the right side, we can move one point to the right using $\sigma_1$ . Starting from $(0,0)$ and following $\sigma_1$ repeatedly, we will reach all the grid points on the positive $x$ -axis.

Now consider the sequence $\sigma_2$ of moves $\langle -6, -1 \rangle, \langle 4, -3 \rangle, \langle 2, 5 \rangle$ . Starting from $(a,b)$ , the path created by $\sigma_2$ is $(a,b), (a - 6,b - 1), (a - 2,b - 4), (a,b + 1).$ If $(a,b)$ is at least 6 units from the left side of the grid, 4 units from the bottom, and 1 unit from the top, we can move one point up using $\sigma_2$ . Starting from $(a,0)$ and following $\sigma_2$ repeatedly, we will reach all the grid points in the first quadrant with $x$ -coordinate equal to $a$ .

Now we have shown that we can reach any point $(a,0)$ on the positive $x$ -axis using $\sigma_1$ , and from $(a,0)$ , we can reach any point $(a,b)$ in the first quadrant using $\sigma_2$ . By symmetry, we can reach any point on the rest of the grid. Therefore, the number of reachable points is $2015^2 = \boxed{4060225}$ .