India — Team Selection Test, 2008

Note that $u = 1$, $v = 2$, $w = 3$ gives $d^3 - 11d - 12 = 0$ which has no integer solutions. Thus $uvw > 8$ and $uv + vw + wu \ge 3(uvw)^{2/3} > 12$. This shows that $d^3 > 12d + 16$ and we infer that $d \ge 5$.

If $d$ is a prime, $d$ has to be an odd prime, say $d = p$. Then $p^3 = p(uv + vw + wu) + 2uvw$, so that $p$ divides $uvw$. Assume $p|u$, and write $u = pu_1$. If $p$ divides $vw$, it must divide either $v$ or $w$. But then
$$p(uv + vw + wu) > pu(v + w) > p^2(p + 1) > p^3.$$
Thus $p$ cannot divide $vw$. The relation now reduces to
$$p^2 = pu_1(v + w) + vw(1 + 2u_1).$$
It follows that $p$ divides $1 + 2u_1$. Thus $1 + 2u_1 \ge p$ or $u_1 \ge (p-1)/2$. Since $v$, $w$ are distinct, we have $v + w \ge 3$. Thus
$$p(uv + vw + wu) > pu(v + w) \ge 3p^2 u_1 \ge \frac{3p^2(p-1)}{2} > p^3,$$
if $p \ge 3$. We conclude that $d$ cannot be an odd prime.

Suppose $d = 6$. Substituting in the given relation, we see that $3|uvw$. Assume $3|u$, so that $u = 3u_1$. We obtain
$$36 = 3u_1(v + w) + vw(1 + u_1).$$
Thus $3|vw(1 + u_1)$. Observe that $3u_1(v + w) < 36$ and $v + w \ge 3$. Hence $u_1 < 4$. Thus either $3|vw$ or $u_1 = 2$. But $u_1 = 2$ shows that $12 = 2(v + w) + vw$ forcing $v = w = 2$. Since $v \ne w$, the other possibility is $3|vw$. Taking $3|v$, we have $v = 3v_1$ and
$$12 = 3u_1v_1 + w(u_1 + v_1) + u_1v_1w.$$
Thus $u_1v_1(3 + w) < 12$. This forces $u_1v_1 < 3$ giving $(u_1, v_1) = (1, 2)$ or $(2, 1)$. But then $u_1 + v_1 = 3$ and $u_1v_1 = 2$, giving $12 = 6 + 5w$. This is impossible for an integer $w$.

We conclude that $d \ge 8$. For $d = 8$, we may take $(u, v, w) = (2, 4, 7)$.