Fix four integers $a,b,c,d$ with $\gcd(a,b)=\gcd(c,d)=1$ and with the condition that $abcd<0$ (this ensures that the two moves are of different sign slope).

Suppose then we have a chess piece that can make moves $(a,b)$, $(-a,-b)$, $(c,d)$, $(-c,-d)$. Place the piece on the bottom left square on an $m\times n$ chessboard. Given an arbitrary number of moves, what is the total number of unique squares the piece can visit?

  • $\begingroup$ You mean for any integer $k$, right? $\endgroup$
    – user123
    Jan 11, 2015 at 23:19
  • $\begingroup$ Yes, thank you. $\endgroup$
    – ruadath
    Jan 11, 2015 at 23:23
  • $\begingroup$ "chess" piece on a "chess" board $\endgroup$ Jan 11, 2015 at 23:25
  • 1
    $\begingroup$ If $\gcd(a, b) = \gcd(c, d) = 1$, then that implies that there is no integer $k$ (except possibly $1$) such that $(a, b) = k(c, d)$. Otherwise $\gcd(a, b) \geq k$. It's enough to say $(a, b) \neq (c, d)$. $\endgroup$
    – Arthur
    Jan 11, 2015 at 23:38


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