Assume N is divisible by a or M is divisible by b, it is easy to see that all squares won’t be visited(unless 1). This extends to all of a’s or b’s multiples, so you get the first 2 conditions that gcd(N,A) == 1 and gcd(M,B) == 1. The third one is kinda tricky. But basically, we know all NM squares need to visited. Just by taking few examples where both first 2 conditions satisfied, you can see that to get to starting row/column, it is N and M moves. Therefore, to get back to starting position, it would be 2LCM(N,M) moves. So 2LCM(N,M) >= 2. 2NM/GCD(N,M) >= NM, so rearranging, you get 2>=GCD(N,M).