$a\mid b^2, b\mid c^2, c\mid a^2\Longrightarrow abc\mid (a+b+c)^k$ Is it true that if $a,b,c$ are positive integers such that $a\mid b^2, b\mid c^2, c\mid a^2$, then $abc$ always divides $(a+b+c)^6$?
If not (I couldn’t find a counterexample), then $abc\mid (a+b+c)^7$ is always true? Or in general, find the minimum value of $k$, such that $abc\mid (a+b+c)^k$.
I proved that $a,b,c\mid (a+b+c)^6$, but unfortunetly that is not enough, since $a,b,c$ are not necessarily relative primes.

If you expand out $(a + b + c)^6$, then all terms are multiples of some $a^ib^jc^k$, with $i+j+k=n$. Now, at least one of $i$, $j$, and $k$ is at least $\frac{n}{3}$. 
Now, we have $abc|a^3b|a^3c^2|a^7$, and symmetrically $abc|b^7$, $abc|c^7$. Thus, we have an upper bound on the solution: $abc|(a+b+c)^{21}$.
To bring that down, we're going to have to do a bit more fiddling with our indices.  
Now, for any $a^ib^jc^k$, if $i, j, k > 0$, then $abc|a^ib^jc^k$, so we just need to worry about the terms where one of them is zero. Since everything's symmetrical, we can just deal with the $k = 0$ case. 
So, we need to know when $abc|a^ib^j$, with $i, j > 0$ (we've already dealt with the case where only one is non-zero above), But notice that we can divide out by an $ab$, so we just need to know when $c|a^{i-1}b^{j-1}$. Since $c|a^2$, $i \geq 3$ suffices. If $i = 2$, then we have $c|a^2|ab^2$, so $i = 2, j \geq 3$ also suffices. If $i = 1$, then we have $c | a^2 | b^4$, so $i = 1, j \geq 5$ suffices. In particular, $i + j \geq 5$ suffices. 
Putting all of those together, whenever $i + j + k \geq 7$, one of the following (or the cyclic permutations of them) holds:  


*

*$i \geq 7$, in which case $abc|a^7|a^ib^jc^k$    

*$i, j, k \geq 1$, in which case $abc|a^ib^jc^k$ since $a|a^i, b|b^j, c|c^k$. 

*$k = 0$, $i \geq 3$, in which case $abc|a^3b|a^ib^jc^k$.  

*$k = 0$, $i = 2$, $j \geq 2$, in which case $abc|a^3b|a^2b^3|a^ib^jc^k$.  

*$k = 0$, $i = 1$, $j \geq 5$, in which case $abc|a^3b|ab^5|a^ib^jc^k$. 


Thus, we have $abc|(a+b+c)^7$. 
Now, we just need to check if $6$ suffices (given your question, I assume you already have a counterexample to show that $5$ does not). 
But notice that, of our options above, 2, 3, 4, and 5 apply to the $i+j+k=6$ case equally, so we know that $abc$ divides all terms of $(a+b+c)^6$, except possibly $a^6$, $b^6$, and $c^6$. We're going to do some symmetry stuff, so we'll start by just considering $a^6$. So we need to know if $bc|a^5$. We know that $bc|a^6$. For any integer $d$ dividing any one of (hence all of) $a, b, c$, say with $x,y,z$ the largest integers such that $p^x|a$,$p^y|b$,$p^z|c$, we have: 
$d^x|a|b^2$, so $x \leq 2y$ (and symmetrically, $y \leq 2z$, $z \leq 2x$). If $g = gcd(b,c)$, then $g^2|bc$ and $g^2|c^2|a^4|a^5$, so $bc|a^5$ if $\frac{bc}{g^2}|\frac{a^5}{g^2}$, so with $b^* = \frac{b}{g}$, $c^* = \frac{c}{g}$, $b^*$ and $c^*$ are coprime, and we wish to know if $b^*c^*|\frac{a^5}{g^2}$, so it suffices to show that $b^*|\frac{a^5}{g^2}$ and $c^*|\frac{a^5}{g^2}$. Multiplying everything by $g^2$, we wish to show that $bg|a^5$ and $cg|a^5$. But $g | c$, so $bg|bc|c^3|a^5$, and $cg|c^2|a^4|a^5$, so indeed $bg|a^5$ and $cg|a^5$, hence $bc|a^5$, hence $abc|a^6$, so indeed, $abc|(a+b+c)^6$. 
