TITLE: Knight move variant: Can it move from $A$ to $B$
QUESTION [7 upvotes]: Given a $N\times N$ "chess" board (Let $N = 10^{100}$) and a knight at
  $(0,0)$. 
Can the knight go to the position $(x, y)$ with jump $[a, b]$ moves
  ? 

jump $[a, b]$ mean: the knight on $(x,y)$ can move to $(x \pm a, y \pm b),\ (x \mp a, y \pm b), \ (x \pm b, y \pm a) \text{ or}\ (x \mp b, y \pm a)$ 

Can I use A* search algorithm to solve this problem ?
Can I solve it without a computer ?

REPLY [4 votes]: To add an algebraic approach going into the same direction as Steven Stadnicki's excellent answer:
Let's call any combination of moves a translation. If the board was infinite, the set of all possible translations of the knight would form a subgroup of the additive group $\Bbb Z \times \Bbb Z$. Hence, the question could be restated as to whether $ (\pm a,b)$,$(\pm b,a)$ generate $\Bbb Z \times \Bbb Z$.
This is obviously false if $(a,b)$ are not coprime. So suppose $a,b$ are coprime and let $$G:=\langle (\pm a,b),(\pm b,a)\rangle \subseteq \Bbb Z \times \Bbb Z$$
It is clear that $(0,2a),(0,2b) \in G$, hence $(0,2) \in G$. Similarly, $(2,0) \in G$, showing $2\Bbb Z \times 2\Bbb Z \subseteq G$.
So the only remaining question is whether the images of $(a,b),(b,a)$ generate $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$. This is the case iff $(a,b) \not \equiv (b,a)$ in $\Bbb Z/2\Bbb Z \times \Bbb Z/2\Bbb Z$, or equivalently $(a,b) \not \equiv (1,1) \bmod 2$.
In other words:

On an infinite board the knight reaches all positions iff $a,b$ are
  coprime and of different parity.

Note that this argument can be translated to $\Bbb N \times \Bbb N$ if you require the board to be bounded to the left and to the bottom of $(0,0)$.