23.24. Consider the lattice shown in Fig. 23.11, and color it in two colors as shown in this figure (white nodes on this figure are not filled; the original square is shaded, with grasshoppers sitting at its white vertices). We will prove that the grasshoppers can only land on white nodes, i.e., under symmetry

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Fig. 23.10

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Fig. 23.11

a white node transitions to a white node. To do this, it is sufficient to prove that under symmetry relative to a white node, a black node transitions to a black node. Let $A$ be a black node, $B$ be a white node, and $A_{1}$ be the image of point $A$ under symmetry relative to $B$. Point $A_{1}$ is a black node if and only if $\overrightarrow{A A_{1}} = 2 m \boldsymbol{e}_{1} + 2 n \boldsymbol{e}_{2}$, where $m$ and $n$ are integers. Clearly, $\overrightarrow{A A_{1}} = 2 \overrightarrow{A B} = 2(m \boldsymbol{e}_{1} + n \boldsymbol{e}_{2})$, so $A_{1}$ is a black node. Therefore, the grasshopper cannot land on the fourth vertex of the square.