South African Mathematical Olympiad, Third Round

We first show that it is not possible for all four polynomials to be non-zero. Suppose they are. Denote the leading coefficients of the polynomials $a(x), b(x), c(x), d(x)$ (which exist, because the polynomials are non-zero) by $A, B, C, D$, respectively. Then the first equation implies $BC = -AD$ and thus $ABCD = -(BC)^2 < 0$. In the second equation, the leading coefficient of $a(x)c(x)$ is $AC$ and the leading coefficient of $(1-x^2)b(x)d(x)$ is $-BD$. Since the degree of $(1-x^2)b(x)d(x)$ is at least 2, these leading coefficients must cancel if we wish to end up with $x+1$ on the right. Thus $AC - BD = 0$, which implies that $ABCD = (AC)^2 > 0$. We have two contradictory inequalities, proving that this system has no solution when all four polynomials are non-zero.

Now suppose that $a(x) = 0$. Then, by the first equation, $b(x) = 0$ or $c(x) = 0$. If $b(x) = 0$, we see that the second equation is not satisfied. If $c(x) = 0$, we see that, again, the second equation is not satisfied, due to differences in degrees. Similarly, when $c(x) = 0$, we find, in a symmetrical way, that there are no solutions.

Henceforth, we assume that both $a(x)$ and $c(x)$ are non-zero. So either $b(x)$ or $d(x)$ (or both) must be zero. From the first equation, $b(x) = 0$ if and only if $d(x) = 0$. It follows that the complete set of solutions is given by all $(a(x), b(x), c(x), d(x))$ where $a(x)c(x) = x+1$ and $b(x) = d(x) = 0$, i.e.,

$$\{(k, 0, \frac{x}{k} + \frac{1}{k}, 0) : k \text{ a non-zero real number}\} \cup \{(\frac{x}{k} + \frac{1}{k}, 0, k, 0) : k \text{ a non-zero real number}\}.$$