
Problem 5. Find all triples of positive integers $(a, b, c)$ such that $a^{3}+b^{3}+c^{3}$ is divisible by $a^{2} b, b^{2} c$ and $c^{2} a$.


Solution: If $d=\operatorname{gcd}(a, b, c)$ it is easy to see that $\left(\frac{a}{d}, \frac{b}{d}, \frac{c}{d}\right)$ is also a solution of the problem. Thus, it suffices to find all triples $(a, b, c)$ such that $\operatorname{gcd}(a, b, c)=1$. Let $\operatorname{gcd}(a, b)=s$ and suppose $s>1$. If $p$ is a prime divisor of $s$ then $p$ divides $a ; p$ divides $b$ and $p$ divides $a^{3}+b^{3}+c^{3}$. Hence $p$ divides $c^{3}$ and so $p$ divides $c$. This is a contradiction to $\operatorname{gcd}(a, b, c)=1$. Therefore $\operatorname{gcd}(a, b)=1$ and by analogy $\operatorname{gcd}(a, c)=\operatorname{gcd}(b, c)=1$. It follows from $a^{2} /\left(a^{3}+b^{3}+\right.$ $\left.c^{3}\right), b^{2} /\left(a^{3}+b^{3}+c^{3}\right), c^{2} /\left(a^{3}+b^{3}+c^{3}\right)$ and $\operatorname{gcd}\left(a^{2}, b^{2}\right)=\operatorname{gcd}\left(a^{2}, c^{2}\right)=$ $\operatorname{gcd}\left(b^{2}, c^{2}\right)=1$ that $a^{3}+b^{3}+c^{3}$ is divisible by $a^{2} b^{2} c^{2}$. In particular $a^{3}+b^{3}+c^{3} \geq a^{2} b^{2} c^{2}$. Wlog we may assume that $a \leq b \leq c$. Then $3 c^{3} \geq a^{3}+b^{3}+c^{3} \geq a^{2} b^{2} c^{2} \Rightarrow c \geq \frac{a^{2} b^{2}}{3}$. Suppose $a>1$. Then $\frac{a^{2} b^{2}}{3}>b^{2} \geq b^{2}+a(a-b)=a^{2}-a b+b^{2}$ and the inequality (1), $c>$ $a^{2}-a b+b^{2}$ holds. From the other hand since $b \geq a \geq 2$ we have that $b^{2} \geq 2 b \geq a+b$ and since $c>b^{2}$ we obtain (2), $c>a+b$. Now (1) and (2) give $\frac{\left(a^{2}-a b+b^{2}\right)(a+b)}{c^{2}}b \geq 3$. Since $1+b^{3}+c^{3} \geq b^{2} c^{2}$, it follows that $2 c^{3}>1+b^{3}+c^{3}$ and so $2 c>b^{2}$ or $c>\frac{b^{2}}{2}$. It follows from $2 c>b^{2}$ that $2 c>b^{2}-b+1$, i.e. (3), $\frac{b^{2}-b+1}{c}\frac{b^{2}}{4}>b+1 \Rightarrow(4) \frac{b+1}{c}<\frac{1}{2}$ hold. Multiplying (3) and (4) gives $\frac{b^{3}+1}{c^{2}}<1$, which is impossible since this number is an integer. Direct verificstion shows that when $b=3$ or $b=4$ we get no new solutions. Therefore all solutions of the problem are triples $(k, k, k)$ and $(k, 2 k, 3 k)$ (and its permutations) for arbitrary positive integer $k$.
