A stable discrete-time LTI system is described by the following difference equation:
$$ y[n] - y[n-1] + Cy[n-2] = x[n] $$
where C is a real number. Determine the range of C so that
(a) the system is causal;
(b) the system is anti-causal;
(c) the system is non-causal (i.e., it has a two-sided impulse response).
It is straight forward to calculate the transfer function:
\begin{align*} Y(z) - z^{-1}Y(z) + Cz^{-2}Y(z) &= X(z) \\ H(Z) = \frac{Y(z)}{X(z)} &= \frac{1}{1 - z^{-1} + Cz^{-2}} \\ \end{align*}
We are given that the system is stable so the ROC must include the unit circle. Therefore there can not be a pole with magnitude \$1\$.
With \$C=0\$, \$H(z)= \frac{1}{1 - z^{-1}}\$, there is a pole at \$1\$ so that is not possible.
With \$C=-2\$, \$H(z)= \frac{1}{1 - z^{-1} - 2z^{-2}} = \frac{\frac{1}{3}}{1+z^{-1}} + \frac{\frac{2}{3}}{1-2z^{-1}}\$, there is a pole at \$-1\$ so that is not possible.
From there, where do I go?
Ultimately this may factor into the form:
$$ \frac{k_1}{1 + a_1 z^{-1}} + \frac{k_2}{1 + a_2 z^{-1}} $$
Is that a causal transfer function or not? The inverse Z transform can yield both a causal and anti-causal impulse response function.