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You can attack this question in a clever, in a general or in a bonus way.

Source is a chess problem of mine (yep, they publish math chess problems in the SCHWALBE, they have a whole category and mine won 1st prize of that year - just showing off).

A Nightrider (N, see next parenthesis) is a Knight (problemists use the German S for that) as a line piece, so it moves e.g. a1-c2-e3-g4 in one move.

Imagine a large chessboard with four squares $A,B,C,D$ (no three on a N line) and a N which can jump from any square to any other, except from $C$ to $D$. You need another $X,Y$-knight (or rider) there, which one? ($6$ solutions if you normalize to $1\leq X\leq Y$)

Clever: A hint: think about necessary angles in the configuration.

General: If you call the tangens values of the lines $AB\dots CD$ (slope against the Cartesian coordinate system) $t_1\dots t_6$, they fulfil a single formula. It's horrible, but at least, if you let all tangens be from $\{t,-t,1/t,-1/t\}$ and solve after $t$, neither of the $6$ possibilities yields a rational solution. And of course you can also solve for any rider combination.

Bonus: This might be too diophantine for Puzzling SE, but so there. What if we generalize the knight to a root knight resp. rider? (So called because the square root $\sqrt{X^2+Y^2}$ is constant. E.g. a $\sqrt{50}$ S can move $5,5$ or $1,7$.) Sorry, we must exclude any "special" configuration of $A,B,C,D$. (E.g. concyclic, otherwise Erdös comes to the rescue. Or a rhombus.)

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  • $\begingroup$ General: It's not THAT horrible and relative quick to prove. Let the counter sides be 16,25,34. Then $t1*t3*t4-t2*t3*t4-t1*t2*t5+t2*t3*t5+t2*t4*t5-t3*t4*t5+t1*t2*t6-t1*t3*t6-t1*t4*t6+t3*t4*t6+t1*t5*t6-t2*t5*t6=0$ $\endgroup$ Commented Jul 14 at 18:33
  • $\begingroup$ Partial answer to bonus: I'm stoopid. Since virtually all of my equations I ever encountered in my recreational math are homogenous, I also assumed these are. They aren't. Which means than any "accidental" rational solution can be scaled. One of the 6 solutions happens to be always rational (regardless of the wurzelrider), the rest of the solution way is routine. $\endgroup$ Commented Jul 19 at 15:06

1 Answer 1

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The six possible values of (X,Y) with X ≤ Y are

(1,62), (2,31), (1,8), (2,11), (38,49), and (13,14).

Explanation:

Let C be the origin. And choose A to be (1,2) without loss of generality (up to rotations and reflections). Note that these positions are all scaled versions of the actual positions on the chess board. Since the problem doesn’t care about the actual positions, this scaled version suffices.

Draw all the lines with slopes ±2 and ±1/2 through C and A. There are six possible locations of B such that an N can move among A, B, and C, but they are not collinear. For each choice of B, there are three possible locations of D such that an N can move between all the four squares except between C and D, and no three points are collinear. All the possible configurations are shown in the figure below. All that’s left to do is to compute the slope of CD. In the figure, I show larger coordinate/smaller coordinate instead of the actual slope and I ignore the signs.

night riders on infinite chessboard

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    $\begingroup$ Noticing that ABC and ABD can only be Pythagorean triangles (otherwise it can't close), you can reduce the case splits a bit, but perfectly legit. $\endgroup$ Commented Jul 22 at 7:39

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