Problem 1376

19. Write $m=[\sqrt[3]{n}]$. To simplify the calculation, we shall assume that $**=1$. Then $a=\sqrt[3]{n}, b=\frac{1}{\sqrt[3]{n}-m}=\frac{1}{n-m^{3}}\left(m^{2}+m \sqrt[3]{n}+\sqrt[3]{n^{2}}\right), c=\frac{1}{b-1}=$ $u+v \sqrt[3]{n}+w \sqrt[3]{n^{2}}$ for certain rational numbers $u, v, w$. Obviously, integers $r, s, t$ with $r a+s b+t c=0$ exist if (and only if) $u=m^{2} w$, i.e., if ( $b-$ 1) $\left(m^{2} w+v \sqrt[3]{n}+w \sqrt[3]{n^{2}}\right)=1$ for some rational $v, w$. When the last equality is expanded and simplified, comparing the coefficients at $1, \sqrt[3]{n}, \sqrt[3]{n^{2}}$ one obtains $$\begin{array}{rlrl} 1: & v+\left(\left(m^{2}+m^{3}-n\right) m^{2}+m\right) w & =n-m^{3}, \\ \sqrt[3]{n}: & \left(m^{2}+m^{3}-n\right) v+ & \left(m^{3}+n\right) w & =0, \\ \sqrt[3]{n^{2}}: & m v+ & \left(2 m^{2}+m^{3}-n\right) w & =0 . \end{array}$$ In order for the system (1) to have a solution $v, w$, we must have $\left(2 m^{2}+m^{3}-n\right)\left(m^{2}+m^{3}-n\right)=m\left(m^{3}+n\right)$. This quadratic equation has solutions $n=m^{3}$ and $n=m^{3}+3 m^{2}+m$. The former is not possible, but the latter gives $a-[a]>\frac{1}{2}$, so $**=1$, and the system (1) in $v, w$ is solvable. Hence every number $n=m^{3}+3 m^{2}+m, m \in \mathbb{N}$, satisfies the condition of the problem.