Solve: $ab+bc+ca\mid (a+b+c)^2$ I couldn't make any progress on this problem, can anyone help?
I found it's the same as:
Find all integers $a,b,c$ such that $ab+bc+ca$ divides $a^2+b^2+c^2$.
I found a solution $a=-b=1$, and $c$ any integer.
Any more solutions?
 A: Yes, there are lots of solutions. These are the ones with $f<g<h\le102$ and $\gcd(f,g,h)=1$.
$$
{1,4,9}\\
{1,9,16}\\
{1,25,36}\\
{1,36,49}\\
{1,49,64}\\
{1,64,81}\\
{1,81,100}\\
{2,3,71}\\
{2,5,71}\\
{3,5,41}\\
{4,9,25}\\
{4,25,49}\\
{4,49,81}\\
{9,16,49}\\
{9,25,64}\\
{9,49,100}\\
{16,25,81}
$$
Mathematica code used:
fmax = 100;
Do[
 If[GCD[f, g, h] != 1, Continue[]];
 If[Mod[f^2 + g^2 + h^2, f g + g h + h f] == 0, Print[{f, g, h}]],
 {f, fmax}, {g, f + 1, fmax + 1}, {h, g + 1, fmax + 2}
 ]

A: Since $(a+b+c)^2 = 2(ab+ac+bc)+(a^2+b^2+c^2)$, if 
$$ (a+b+c)^2 = k(ab+ac+bc), \tag{1}$$
then $k\geq 2$, and for $k=2$ we have only the trivial solution $(a,b,c)=(0,0,0)$. 
Assuming $k=3$, we have:
$$ a^2+b^2+c^2 = ab+ac+bc \tag{2}$$
and by the Cauchy-Schwarz inequality $(2)$ holds only for $a=b=c$.
Assuming $k=4$ we have the parametric solution:
$$ (a,b,c) = (m^2,n^2,(n+m)^2) \tag{3}$$
so there are plenty of solutions, and even more can be computed by Vieta jumping. 
Markov triples are deeply related.
A: For such equations:
$$(a+b+c)^2=-t^2(ab+ac+bc)$$
$t$ - you can specify any, then decisions can be recorded.
$$a=p^2-2(t^2+t+2)ps+(2t^3+t^2+4t+4)s^2$$
$$b=-p^2+2(t^2-t+2)ps+(2t^3-t^2+4t-4)s^2$$
$$c=t(p^2-(t^2+4)s^2)$$
$p,s$ - integers asked us.
