Question

Let $a,b,c$ be positive integers such that $a$ divides $b^5$, $b$ divides $c^5$, and $c$ divides $a^5$. Prove that $abc$ divides $(a+b+c{)}^{31}$.

Answer 1

If a prime p divides a, then $p\mid b^5$ and hence $p\mid b$.
This implies that $p\mid c^4$ and hence $p\mid c$.
Thus every prime dividing a also divides b and c. By symmetry, this is true for b and c as well.
$\therefore a,b,c$ have the same set of prime divisors.
Let $p^x\parallel a,p^y\parallel b$ and $p^z\parallel c$. (Here we write $p^x\parallel a$ to mean $p^x\mid a$ ) We may assume min {x, y, z} = x. Now $b\mid c^5$ implies that $y\le 5z;c\mid a^5$ implies that $z\le 5x$.
We obtain $y\le 5z\le 25x$.
Thus $x+y+z$ $\le$ $x+5x+25x=31x.$
Hence the maximum power of p that divides $abc$ is $x+y+z$ $\le$ $31x$.
Since x is the minimum among $x,y,z,$ $p^x$ divides $a,b,c$.
$\therefore p^x$ divides $a+b+c$.
$\to p^{31x}$ divides $(a+b+c{)}^{21}$.
Since $x+y+z$ $\le$ $31x$, it follows that $px+y+z$ divides $(a+b+c{)}^{31}$.
This is true of any prime $p$ dividing $a,b,c.$
$\therefore abc$ divides $(a+b+c{)}^{31}$.