Solution suggested by the student Mahdi Al-Shaikh Saleh. Consider the lattice $\mathbb{Z}^{2}$ and $(x_{a}, y_{a}),(x_{b}, y_{b})$ the coordinates of positions of pebbles $A, B$ respectively. If we relocate one of the pebbles, say $B$, at a new position, $(x_{b}^{\prime}, y_{b}^{\prime})$, because $(x_{b}-x_{a})^{2}+(y_{b}-y_{a})^{2}=(x_{b}^{\prime}-x_{a})^{2}+(y_{b}^{\prime}-y_{a})^{2}$, by considering this equality in modulo $4$, we deduce that the number of even terms among $x_{b}-x_{a}, y_{b}-y_{a}$ is preserved after each relocation. Whenever both terms are even, one can cancel the common $2$ by considering a sublattice to end up with two terms with at most one even term. In this way, we can consider two cases:

- If only one term among $x_{b}-x_{a}, y_{b}-y_{a}$ is odd, we color the vertices of the lattice in blue and red by alternating the colors between each two consecutive vertices in a horizontal line and two consecutive vertices in a vertical line.

- If both terms $x_{b}-x_{a}, y_{b}-y_{a}$ are odd, we color the vertices of the lattice in blue and red by making the colors of all vertices in a same horizontal line the same and alternating the colors between each two consecutive horizontal lines.

In both cases, the colors of the vertex positions of pebbles $A$ and $B$ are different, and at each relocation, the colors of their vertex new positions remains unchanged. So, in both cases, it is impossible after finite number of steps to switch positions of the pebbles $A$ and $B$.