<s>[INST] 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?
 [/INST] 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.
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