The official solution came out today:
SOLUTION: The first thing to do here is think of the initial square as a cell of the plane grid, e.g., the points (0,0), (0,1), (1,0) and (1,1) on the X-Y plane. Then the pegs will always be on grid points.
Grid points, however, have four possible parities: each coordinate can be even or odd. When a peg jumps, its parity is preserved; its X-coordinate goes up or down by an even number, and likewise its Y-coordinate.
The points of the unit cell above have all four parities, so the corner that starts without a peg can never be occupied.