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There are only finitely many such matrices. Let us consder the circular matrix say $A$, with the first row $(i,j,k,m)$. Then $det(A)=FGH$, where $$ F=i-j+k-m,\; G=i+j+k+m,\;H=(i-k)^2+(j-m)^2 $$ We need to have $det(A)=\pm 1$. Thus, we need to play with the following system of equations \begin{align*} (A) \quad &F=G=H=1,\\ (B) \quad &F=G=-1, H=1,\\ (C) \quad &F=-1, G=1, H=1,\\ (D) \quad &F=1, G=-1, H=1. \end{align*} Is is a simple excercise to prove that each of the above systems has only finitely many solotions (under the condition that $i, j, k, m\in\mathbb{Z}$).