To see this, select first a random spin position \( x,y \) and assume that this spin and its nearest neighbors are all pointing up. The energy for this configuration is \( E=-4J \). Now we flip this spin as shown below. The energy of the new configuration is \( E=4J \), yielding \( \Delta E=8J \). $$ \begin{equation*} E=-4J\hspace{1cm}\begin{array}{ccc} & \uparrow & \\ \uparrow & \uparrow & \uparrow\\ & \uparrow & \end{array} \hspace{1cm}\Longrightarrow\hspace{1cm} E=4J\hspace{1cm}\begin{array}{ccc} & \uparrow & \\ \uparrow & \downarrow & \uparrow\\ & \uparrow & \end{array} \end{equation*} $$ The four other possibilities are as follows $$ \begin{equation*} E=-2J\hspace{1cm}\begin{array}{ccc} & \uparrow & \\ \downarrow & \uparrow & \uparrow\\ & \uparrow & \end{array} \hspace{1cm}\Longrightarrow\hspace{1cm} E=2J\hspace{1cm}\begin{array}{ccc} & \uparrow & \\ \downarrow & \downarrow & \uparrow\\ & \uparrow & \end{array} \end{equation*} $$ with \( \Delta E=4J \), $$ \begin{equation*} E=0\hspace{1cm}\begin{array}{ccc} & \uparrow & \\ \downarrow & \uparrow & \uparrow\\ & \downarrow & \end{array} \hspace{1cm}\Longrightarrow\hspace{1cm} E=0\hspace{1cm}\begin{array}{ccc} & \uparrow & \\ \downarrow & \downarrow & \uparrow\\ & \downarrow & \end{array} \end{equation*} $$ with \( \Delta E=0 \), $$ \begin{equation*} E=2J\hspace{1cm}\begin{array}{ccc} & \downarrow & \\ \downarrow & \uparrow & \uparrow\\ & \downarrow & \end{array} \hspace{1cm}\Longrightarrow\hspace{1cm} E=-2J\hspace{1cm}\begin{array}{ccc} & \downarrow & \\ \downarrow & \downarrow & \uparrow\\ & \downarrow & \end{array} \end{equation*} $$ with \( \Delta E=-4J \) and finally $$ \begin{equation*} E=4J\hspace{1cm}\begin{array}{ccc} & \downarrow & \\ \downarrow & \uparrow & \downarrow\\ & \downarrow & \end{array} \hspace{1cm}\Longrightarrow\hspace{1cm} E=-4J\hspace{1cm}\begin{array}{ccc} & \downarrow & \\ \downarrow & \downarrow & \downarrow\\ & \downarrow & \end{array} \end{equation*} $$ with \( \Delta E=-8J \).