The Knight's Tour problem consists of finding a Hamiltonian path for the knight on a given set of points so that the knight can visit exactly once every vertex of the mentioned set. In the present paper we provide a $5$-dimensional counterexample to the well-known statement that it is not ever possible for a knight to visit once every vertex of $C(3,k):=\{\underbrace{\{0,1,2\} \times \{0,1,2\}\times \cdots \times \{0,1,2\}}_\textrm{k-times}\}$ by performing a sequence of $3^k-1$ jumps of standard length, since the most accurate answer to the original question actually depends on which mathematical assumptions we are making at the beginning of the game, when we decide to extend a planar chess piece to the third dimension and above. Our counterintuitive outcome follows from the observation that we can alternatively define a $2$D knight as a piece that moves from one square to another on the chessboard by covering a fixed Euclidean distance of $\sqrt{5}$ so that also the statement of Theorem 3 in [Erde, J., Gol\'enia, B., & Gol\'enia, S. (2012), The closed knight tour problem in higher dimensions, The Electronic Journal of Combinatorics, 19(4), \#P9] does not hold anymore for such a Euclidean knight, as long as a $2 \times 2 \times \cdots \times 2$ chessboard with at least $2^7$ cells is given. Moreover, we construct a classical closed knight's tour on $C(3,4)-\{(1,1,1,1)\}$ whose arrival is at a distance of $2$ from $(1,1,1,1)$, and then we show a closed Euclidean knight's tour on $\{\{0,1\}\times\{0,1\}\times\{0,1\}\times\{0,1\}\times\{0,1\}\times\{0,1\}\times\{0,1\}\}\subseteq \mathbb{Z}^7$.
[https://arxiv.org/abs/2309.09639v1]