<s>[INST] The sum of the least common multiple and greatest common divisors of the product 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 $p^2=y^2+yz+z^2$?
 [/INST] 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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