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This question already has an answer here:

Consider an infinite chess board. A knight moves 2 squares forward on one direction, then turn left or right, move 1 square further on. Let's denote this a normal knight, or $\langle 2,1\rangle$ knight. And it's known that a knight could reach any square on the board, i.e. denoting the starting square as $(0,0)$, a $\langle 2,1\rangle$ knight can arrive any square $(m,n)$.

Now let's say this knight gets drunk and started to walk $\langle p,q\rangle$...

Of course, if both $p$ and $q$ are even numbers, a $\langle p,q\rangle$ knight won't arrive $(m,n)$ if both $m$ and $n$ are odd numbers.

Then brings to the question, in which situation, a drunk $\langle p,q\rangle$ knight still could reach any square $(m,n)$ ?

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marked as duplicate by user10354138, Ingix, YuiTo Cheng, Yanior Weg, Peter Jun 22 at 8:55

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