India — Team Selection Test, 2010

Let $\alpha$, $\beta$, $\gamma$ be the roots of the cubic. Then we have
$$\alpha + \beta + \gamma = -a, \quad \alpha\beta + \beta\gamma + \gamma\alpha = b, \quad \alpha\beta\gamma = -c.$$
Thus $\alpha^2 = \beta + \gamma = -a - \alpha$. We also observe that $\alpha \neq 0$; otherwise $c = 0$ contradicting that $c$ is odd. Hence $\beta\gamma = -c/\alpha$ and
$$b = \alpha(\beta + \gamma) + \beta\gamma = \alpha^3 - \frac{c}{\alpha},$$
or $\alpha^4 - b\alpha - c = 0$. This gives $(\alpha + \alpha)^2 - b\alpha - c = 0$, which simplifies to $a(2\alpha - b - 1) = c + a(a - 1)$. Note that the right side is odd and hence is not equal to 0. We thus get
$$\alpha = \frac{c + a(a - 1)}{2a - b - 1}.$$
This shows that $\alpha$ is a rational number. Now $\alpha$ is a rational root of a monic polynomial with integer coefficients. Hence it must be an integer. Since $a = -\alpha(\alpha + 1)$, $a$ is even.
Suppose $\beta = \gamma$. Then $2\beta = \beta + \gamma = -a - \alpha$, so that $\beta$ is a rational number. Hence $\beta(= \gamma)$ is an integer. Again $2\beta = -a - \alpha$ shows that $\alpha$ is even. But then $c = -\alpha\beta\gamma$ shows that $c$ is even. Hence $\beta = \gamma$ is not possible.