Question

Let $a,b,c$ be positive integers such that $a$ divides $b^3$, $b$ divides $c^3$, and $c$ divides $a^3$. Prove that $abc$ divides $(a+b+c{)}^{13}$.

Answer 1

If a prime p divides a, then $p|b^3$ and hence p | b. This implies that $p|c^3$ and hence p | c. Thus every prime dividing a also divides b and c. By symmetry, this is true for b and c as well. We conclude that a, b, c have the same set of prime divisors.
Let $p^x||a,p^y||b$ and $p^z||c$. (Here we write $p^x||a$ to mean $p^x|a$ and $p^{x+1}|/a$.) We may assume min {x, y, z} = x. Now $b|c^3$ implies that $y\le 3z;c|a^3$ implies that $z\le 3x$. We obtain
Thus $x+y+z\le x+3x+9x=13x$. Hence the maximum power of p that divides abc is $x+y+z\le 13x$. Since x is the minimum among x, y, z, whence $p^x$ divides each of a, b, c. Hence $p^x$ divides a + b + c. This implies that $p^{13x}$ divides $(a+b+c{)}^{13}$. Since $x+y+z\le 13x$, it follows that $p^{x+y+z}$ divides $(a+b+c{)}^{13}$. This is true of any prime p dividing a, b, c. Hence abc divides $(a+b+c{)}^{13}$.