A bunny is hopping around the plane. She must

  1. land on the following $25$ points on the main diagonal (multiples of 4): $(4, 4)$, $(8, 8)$, $(12, 12)$, $\ldots,$ $(100, 100)$, (The bunny can land on other points too, but the goal is to hit these multiples of four.)

  2. only land on lattice points $\mathbb{Z}^2$,

  3. only jump integer distances, and

  4. move so no pair of jumps are parallel or perpendicular to each other. That means if the bunny jumps directly north for its first jump, it cannot jump directly north, south, east or west for any of the remaining jumps.

The bunny can start by jumping to any lattice point you wish. How can the bunny accomplish this task?

  • $\begingroup$ Any of the remaining jumps or just the next jump? Say the first jump is north, and the second jump is south east, can the third jump be north again? Are you only considering consecutive pairs of jumps, or any two in the whole set of jumps? $\endgroup$
    – CodeNewbie
    Commented Jul 10, 2015 at 8:11
  • $\begingroup$ @CodeNewbie: Any two jumps in the whole set. That is, the third jump cannot be north again. $\endgroup$ Commented Jul 10, 2015 at 14:55
  • $\begingroup$ I'm surprised this question doesn't have more upvotes. It's a neat problem. $\endgroup$
    – COTO
    Commented Jul 10, 2015 at 16:47
  • $\begingroup$ @COTO: Thanks. In retrospect I think it had a few too many arbitrary seeming conditions, and even so I didn't quite restrict the solution space quite as much as I wanted ... $\endgroup$ Commented Jul 11, 2015 at 3:50

1 Answer 1


If $(a,b,c)$ is a Pythagorean triple, the bunny can do moves $(a,b)$ followed by $(-a,b)$ to move a displacement of $2b$ in any orthogonal direction.

Using triples of the form $(m^2-1,2m,m^2+1)$, we can obtain any even number $b=2m$, and so any orthogonal displacement $4m$. By moving $4(m+1)$ spaces, then $4m$ spaces the other way, we can move $4$ spaces.

With $m$ sufficiently large, we can guarantee that we haven't used $m$ and $m+1$ before, so this move isn't parallel or perpendicular to any previous move. (When $m$ is odd, the triple has a GCD of 2, so we have to care that its halved version hasn't been used either.)

The bunny can start at $(0,0)$ and repeatedly move $4$ steps and $4$ steps right to hit every required point.

  • 2
    $\begingroup$ I think this is a good answer, but it would be even nicer, if you'd give the "jump" examples for the first couple of steps as illustration. $\endgroup$
    – BmyGuest
    Commented Jul 10, 2015 at 9:06

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