A graph problem. Two friends $A$ & $B$ are initially at points $(0,0)$ & $(12,7)$ respectively on the infinite grid plane. $A$ takes steps of size $4$ units and $B$ takes steps of size $6$ unit along the grid lines. Show that it is not possible for them to meet at a point.
It is my problem.I can find the number of ways to go from $A$ to $B$, but! how I prove or disprove it. Thank you.

I assume that the definition of "meet" by the OP is that both $A$ and $B$ must end a move at the same point.
Consider a move that $A$ makes from $(x_1,y_1)$ to $(x_2,y_2)$
Notice that the sum of the "net movement" of $A$ in the $x$ and $y$ directions (i.e. $|x_1 - x_2| + |y_1 - y_2|$ where $|x|$ is the absolute value of $x$) is always even. We can prove this by considering moves in opposite directions. Notice that a move in one direction cancels a move in the opposite direction, and that this "cancellation" occurs in pairs of moves. Thus the net movement (i.e. the non-cancelled moves) must have a sum equal to $total \space moves - cancelled \space moves$. Since both numbers are even, the sum of the net movement is  even. Similarly for $B$.
Thus if we sum the $x$ and $y$ coordinate for any where $A$ can be, it must have the same parity as the sum of $A$'s $x$ and $y$ coordinate at the start since all moves will have a sum of net movement of an even number, preserving the parity. Similarly for $B$.
Since in order for them to meet, the sum of the $x$ and $y$ coordinate of the meeting square has to be the same for both, thus the parity has to be the same.
However, the sum of the $x$ and $y$ coordinates of $(0,0)$ is an even number, while the sum of the $x$ and $y$ coordinates of $(12,7)$ is odd.
This implies that they cannot land on a square together since one of the parities must change, which is disallowed by the moves.
