<s>[INST] show that $ab+bc+ca|(b^2-ac)^2$ Let $a,b,c$ be postive integers,and such $(a,b)=(b,c)=(c,a)=1$,if 
$$ab+bc+ca|(a^2-bc)^2$$
show that
$$ab+bc+ca|(b^2-ac)^2$$
Own idea,maybe we look for some  identities?
 [/INST] $$0\equiv (a^2-bc)^2 \equiv (a^2-(-ab-ac))^2 = a^2(a+b+c)^2 \bmod (ab+bc+ca).$$
However, $\gcd(a,bc)=1 \implies \gcd(a,ab+bc+ca)=1,$ so 
$$ab+bc+ca | (a+b+c)^2.$$
$$ab+bc+ca | (b^2+ab+bc)^2.$$
$$ab+bc+ca | (b^2-ca)^2.$$
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