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Find number of solutions in positive integers the equation $$[a^2,b^2]+[b^2,c^2]+[c^2,a^2]=(a^2,b^2)(b^2,c^2)(c^2,a^2),$$ where $[m,n] -$ least common multiple, $(m,n) -$ greatest common divisors of positive integers $m$ and $n$.

My work so far:

Let $a=px, b=py, c=pz$, where $p,x,y,z -$ pairwise relatively prime numbers. Then $$(pxy)^2+(pyz)^2+(pzx)^2=(p^2)^3$$ $$x^2y^2+y^2z^2+z^2x^2=p^4$$ $$x^2(y^2+z^2)=(p^2-yz)(p^2+yz)$$ If $p^2=y^2+yz+z^2$, then $$x^2(y^2+z^2)=(y^2+2yz+z^2)(y^2+z^2)$$ $$x=y+z$$ Under such conditions, we have a solution $$(p(y+z);py,pz).$$

Answer: Can we show that there are infinitely many triples of positive integers $(p,y,z)$ for which the equality $p^2=y^2+yz+z^2$?

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The first step isn't possible: $p$ must be the gcd, and you cannot then require that $x$, $y$, and $z$ be coprime, e.g., if $a=6$, $b=10$, $c=15$.

If you write $a$ as $dABC$, where $d$ is $gcd(a,b,c)$, $A$ is $gcd(a/d,b/d)$, and $B$ is $gcd(a/d,c/d)$, and $b$ and $c$ in similar fashion, you should be able to cancel things, and there don't seem to be any solutions.

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    $\begingroup$ Actually, there are solutions. $\endgroup$ – Davey Feb 17 '16 at 15:34
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    $\begingroup$ There are either none or infinitely many, as $C$ and the corresponding values for b and c cancel. $\endgroup$ – Davey Feb 17 '16 at 15:40

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