Question

Let a, b, c be positive integers such that a divides $b^4$, b divides $c^4$ and c divides $a^4$. Prove that abc divides $(a+b+c{)}^{21}$

Answer 1

If a prime p divides a, then $p|b^4$ and hence p | b.
This implies that $p|c^4$ 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, $b|c^4$ implies that $y\le 4z;c|a^4$ implies that $z\le 4x$. We obtain $y\le 4z\le 16x$
Thus $x+y+z\le x+4x+16x=21x$.
Hence the maximum power of p that divides abc is $x+y+z\le 21x$.
Since x is the minimum among $x,y,z,p^x$ divides a, b, c.
$\therefore p^x$ divides a + b + c.
$\to p^{21x}$ divides $(a+bc{)}^{21}$. Since $x+y+z\le 21x$, it follows that $p^{x+y+z}$ divides $(a+b+c{)}^{21}$.
This is true of any prime p dividing a, b, c.
$\therefore abc$ divides $(a+b+c{)}^{21}$