We will show that the only such number is $a=0$. Let $A=\left(\begin{array}{cccc} a & -a & -1 & 0 \\ a & -a & 0 & -1 \\ 1 & 0 & a & -a \\ 0 & 1 & a & -a \end{array}\right)$ and suppose that $A=B^{2}$. It is easy to compute the characteristic polynomial of $A$, which is $p_{A}(x)=\operatorname{det}(A-x I)=\left(x^{2}+1\right)^{2}$. By the Cayley-Hamilton theorem we have $p_{A}\left(B^{2}\right)=p_{A}(A)=0$. Let $\mu_{B}(x)$ be the minimal polynomial of $B$. The minimal polynomial divides all polynomials that vanish at $B$; in particular $\mu_{B}(x)$ must be a divisor of the polynomial $p_{A}\left(x^{2}\right)=\left(x^{4}+1\right)^{2}$. The polynomial $\mu_{B}(x)$ has rational coefficients and degree at most 4. On the other hand, the polynomial $x^{4}+1$, being the 8th cyclotomic polynomial, is irreducible in $\mathbb{Q}[x]$. Hence the only possibility for $\mu_{B}$ is $\mu_{B}(x)=x^{4}+1$. Therefore, $A^{2}+I=\mu_{B}(B)=0$. Since we have $A^{2}+I=\left(\begin{array}{cccc} 0 & 0 & -2 a & 2 a \\ 0 & 0 & -2 a & 2 a \\ 2 a & -2 a & 0 & 0 \\ 2 a & -2 a & 0 & 0 \end{array}\right)$ the relation forces $a=0$. In case $a=0$ we have $A=\left(\begin{array}{cccc} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right)=\left(\begin{array}{cccc} 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right)^{2}$ hence $a=0$ satisfies the condition.