IMC 2006 · Day 2 · P11

Substituting $y = x + m$, we can replace the equation by $$y^3 - ny + mn = 0.$$ Let two roots be $u$ and $v$; the third one must be $w = -(u + v)$ since the sum is 0. The roots must also satisfy $$uv + uw + vw = -(u^2 + uv + v^2) = -n, \quad \text{i.e.} \quad u^2 + uv + v^2 = n$$ and $$uvw = -uv(u + v) = mn.$$ So we need some integer pairs $(u, v)$ such that $uv(u + v)$ is divisible by $u^2 + uv + v^2$. Look for such pairs in the form $u = kp$, $v = kq$. Then $$u^2 + uv + v^2 = k^2 (p^2 + pq + q^2),$$ and $$uv(u + v) = k^3 pq (p + q).$$ Chosing $p, q$ such that they are coprime then setting $k = p^2 + pq + q^2$ we have $\dfrac{uv(u+v)}{u^2 + uv + v^2} = p^2 + pq + q^2$.

Substituting back to the original quantites, we obtain the family of cases $$n = (p^2 + pq + q^2)^3, \qquad m = p^2 q + p q^2,$$ and the three roots are $$x_1 = p^3, \qquad x_2 = q^3, \qquad x_3 = -(p+q)^3.$$