Is a Difference Equation causal, anti causal, or non causal?
This is a root locus analysis with variable K. 

The poles are:
$$
p_{1,2}=\frac{1}{2}(1 \pm (1 - 4c)^{1/2}), |p_2|&lt;|p_1|
$$

For \$c&lt;=1/4\$, the poles are real. \$p_{1}=1\$ when \$c = 0\$.

For \$c&gt;1/4\$, the poles are complex, and their magnitude is:
$$
|p_{1,2}|=\sqrt{\frac{1}{4}(1 + (4c-1))}=\sqrt{c}
$$
which is equal to 1 when \$c=1\$. 

Hence, for \$c\$, \$p_1\$ and \$p_2\$:


\$c=0: p_1=1\$ (critical), \$p_2=0\$ (stable)
\$c=1: p_{1,2}=1/2 \pm i\sqrt(3)/2\$ (underdamped stable,\$&lt;1\$)
\$c=1/4: p_1=1/2, p_2=1/2\$ (overdamped stable,\$&lt;1\$)


Because the time response is given, with negative delays, the system is casual, independently of \$c\$. 

If only \$H(z)\$ were given and no time response, we obtain the explicit division, for inspect each converging term involving infinite sums:
$$
  H(Z) = \frac{1}{p_1-p_2}(\frac{p_1}{1 - p_1 z^{-1}}-\frac{p_2}{1 - p_2 z^{-1}}) 
$$

Hence:

For a Causal System (no diverging components on the sum, for any \$c\$): 

$$
|p_1 z^{-1}|&lt;1 and |p_2 z^{-1}|&lt;1  \rightarrow  max(|p_1|,|p_2|)&lt;|z|  
$$

if \$c&lt;=1/4\$: \$\frac{1}{2}(1 + (1 - 4c)^{1/2})&lt;|z|\$ 

if \$c&gt;1/4\$: \$\sqrt{c}&lt;|z|\$  

For a Anti-Causal System (both components diverge): 

$$
|p_1 z^{-1}|&gt;1 and |p_2 z^{-1}|&gt;1  \rightarrow  min(|p_1|,|p_2|)&gt;|z|
$$

if \$c&lt;=1/4\$: \$\frac{1}{2}(1 - (1 - 4c)^{1/2})&gt;|z|\$ 

if \$c&gt;1/4\$: \$\sqrt{c}&gt;|z|\$ 

For a Non-Causal System (one component diverge and the other converge): 

$$
|p_1 z^{-1}|&gt;1 and |p_2 z^{-1}|&lt;1  \rightarrow  |p_2|&lt;|z|&lt;|p_1|
$$

if \$c&lt;=1/4\$: \$\frac{1}{2}(1 - (1 - 4c)^{1/2})&lt;|z|&lt;\frac{1}{2}(1 + (1 - 4c)^{1/2})\$

if \$c&gt;1/4\$: \$\sqrt{c}=|z|\$ 

in MATLAB, making:

h = tf([1],[1 -1 0],1); rlocus(h); axis equal;


you obtain the 'variable locus', though the correct interpretation of the c values is artificial here.



Q.E.P.D.