Indija TS 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 $u=p{u}_1$. 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 $$p^2=p{u}_1(v+w)+vw(1+2u_1).$$ 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 $$12=3u_1{v}_1+w(u_1+v_1)+u_1{v}_{1}w.$$ Thus $u_1{v}_1(3+w)<12$. This forces $u_1{v}_1<3$ giving $(u_1,v_1)=(1,2)$ or $(2,1)$. But then $u_1+v_1=3$ and $u_1{v}_1=2$, giving $12=6+5w$. This is impossible for an integer $w$.

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