India — Team Selection Test, 2009

Observe that $\prod_{j=1}^{n} a_j = 1$. Hence by dividing by this product, we may write
$$f(x) - g(y) = \prod_{j=1}^{n} x - \alpha_j y + \beta_j.$$
We observe that
$$\prod_{j=1}^{n} (x - \alpha_j y) = x^n - y^n = \prod_{j=1}^{n} (x - w^j y),$$
where $w$ is the primitive $d$-th root of unity. Hence, after renumbering if necessary, we may write $\alpha_j = w^j$, $1 \le j \le n$. Thus we have
$$f(x) - g(y) = \prod_{j=1}^{n} (x - w^j y + \beta_j).$$
Define
$$a = \frac{\beta_1 - w\beta_n}{w - 1}, \quad b = \frac{\beta_1 - \beta_n}{w - 1}.$$
Consider $F(x) = f(x + a)$, $G(y) = g(y + b)$. It is easy to check that
$$F(x) - G(y) = (x - y)(x - wy) \prod_{j=2}^{n-1} (x - w^j y + \gamma_j),$$
where $\gamma_j = \beta_j + a - w^j b$, $2 \le j \le n-1$. Putting $y = x$, we see that $F(x) = G(x)$. Putting $x = wy$, we also see that $F(wy) = G(y) = F(y)$. Thus whenever $\alpha$ is a root of $F(x) = 0$, so is $w\alpha$. Thus $F(y) = 0$ has roots $\alpha, w\alpha, \dots, w^{n-1}\alpha$. We thus obtain
$$F(y) = G(y) = (y - \alpha)(y - w\alpha) \cdots (y - w^{n-1}\alpha) = y^n - \alpha^n.$$
This gives
$$f(x) = F(x - a) = (x - a)^n - \alpha^n, \quad g(y) = (y - b)^n - \alpha^n.$$