Total bugs => 65
total squares => 81
If we take 64 bugs, then we can arrange them together into a matrix of 8 × 8 square, so their is a possibility that No 2 bugs are in same square, because we can move all the bugs vertically upward in 1st step, then Horizontally left in 2nd step vertically down in the third step, and in the 4th step horizontaly left and so on.
But , If we take 65 bugs so one horizontal or vertical row of square will fill with bugs. So we can not perform the above process in this situation [due to extra 65th bug] so after some move their will be 2 bugs in same square.
“OR”
We have consider 16 shaded squares. Let we have a bug in the shaded Square. So in at most 4 moves, Bug will be in any shaded square again.
And if we have a bug in the un-shaded square, in at most 3 moves, bug will be in any shaded square again So, if we have total 65 bugs in these 81 squares, some of them will be in shaded square and some of them in un-shaded square.
So after 3 or 4 moves all the bugs need to be in shaded square. So their will exist atleast one move in which 2 bugs will get into the same shaded square-
