Harvard-MIT Mathematics Tournament, Team Round B

Solution:
That $0$ is a possible value of $a + d$ can be seen by taking $m(x) = -x$, i.e., $a = -d = 1$, $b = c = 0$. We will now show that $0$ is the only possible value of $a + d$.

The equation $x = m(m(x))$ implies
$$x = \frac{\left(a^{2} + b c\right) x + (a + d) b}{(a + d) c x + \left(b c + d^{2}\right)}$$
which in turn implies
$$(a + d)\left[c x^{2} + (-a + d) x - b\right] = 0$$
Suppose for the sake of contradiction that $a + d \neq 0$. Then the above equation would further imply
$$c x^{2} + (-a + d) x - b = 0, \quad x(c x + d) = a x + b,$$
which would imply $x = m(x)$ for any $x$ except possibly $-d / c$. But of course $-d / c$ is not a root of $x = m(m(x))$ anyway, so in this case, all solutions of $x = m(m(x))$ are also solutions of $x = m(x)$, a contradiction. So our assumption was wrong, and in fact $a + d = 0$, as claimed.