Question

Let $a,b,c$ be positive integers such that a divides $b^2$, b divides $c^2$ and c divides $a^2$. Prove that $abc$ divides $(a+b+c{)}^7$

Answer 1

If a prime p divides a, then $p|b^2$ and hence $p|b$. This implies that $p|c^2$ 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 $pz||c$. (Here we write $p^x||a$ to mean $px|a$ and $p^{x+1}|/a.)$
Assume min {x, y, z} = x.
Now $b|c^2$ implies that $y\le 2z$; $c|a^2$ implies that $z\le 2x$.
We obtain $y\le 2z\le 4x$.
$\therefore x+y+z\le x+2x+4x=7x$.
Hence the maximum power of p that divides abc is $x+y+z\le 7x$.
Since x is the minimum among x, y, z, $p^x$ divides a, b, c.
$\therefore p^x$ divides a + b + c.
This implies that $p^{7x}$ divides $(a+b+c{)}^7$.
Since $x+y+z\le 7x$, it follows that $p^{x+y+z}$ divides $(a+b+c{)}^7$.
This is true of any prime p dividing a, b, c. Hence abc divides $(a+b+c{)}^7$.