Gauss and matrices

We want $\left(\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right) \; = \; A^2 \; = \; \left(\begin{array}{cc} 5 & a \\ b & c \end{array}\right)^2 \; = \; \left(\begin{array}{cc}25 + ab & a(5+c) \\ b(5+c) & c^2 + ab \end{array}\right)$ so we must have $c^2 = 25$ , $ab=-26$ and $a(5+c) = b(5+c) = 0$ .

If $c=5$ we would see that $a=b=0$ , and so the equations are inconsistent. Thus it follows that $c=-5$ , and so the only other condition required is $ab=-26$ . Thus the number of matrices of the desired form is simply equal to the number of divisors of $-26$ in $\mathbb{Z}[i]$ . Now $26 \; = \; (1+i)(1-i)(2+3i)(2-3i) = i(1-i)^2(2+3i)(2-3i)$ is a factorization of $26$ into distinct irreducibles and units of $\mathbb{Z}[i]$ . Thus the most general form of $a$ is $a \; = \; u(1-i)^p(2+3i)^q(2-3i)^r$ where $u \in \mathbb{Z}[i]$ is a unit and $p,q,r$ are integers with $0 \le p \le 2$ , $0 \le q,r \le 1$ . Since there are $4$ units in $\mathbb{Z}[i]$ (namely $1,-1,i,-i$ ), we deduce that there are $4\times 3 \times2\times2 = 48$ divisors of $-26$ , and hence $\boxed{48}$ matrices.