When $\gcd(a,b,c)\cdot \text{lcm}(a,b,c)=\sqrt{abc}$ Recently, I have found this problem:

Given three integer numbers $a,b,c$ such that $1\leq a,b,c\leq 30$ and the following relation holds:
$$\gcd(a,b,c)\cdot \text{lcm}(a,b,c)=\sqrt{abc}$$
How many different tuples $(a,b,c)$ are there?

To solve this, I thought to write:
$$\text{lcm}(a,b)\cdot c=\gcd(\text{lcm}(a,b),c)\cdot \text{lcm}(\text{lcm}(a,b),c)$$
And:
$$\gcd(a,b)\cdot c=\gcd(\gcd(a,b),c)\cdot \text{lcm}(\gcd(a,b),c)$$
So, I have:
$$\frac{ab\cdot c^2}{\gcd(\text{lcm}(a,b),c)\cdot \text{lcm}( (\gcd(a,b),c)}=\sqrt{abc}$$
But here I am stuck. Any idea of how to proceed?
Thank you.
 A: Suppose $(a,b,c)$ is a solution with $abc\ne 0$ and $a\le b\le c$, and suppose $p$ is a prime with $p^r||a$, $p^s||b$, $p^t||c$ ($||$ means that the quotient is not divisible by $p$). After renaming $r$, $s$, $t$ we may assume $r\le s\le t$. Then the power of $p$ in $\sqrt{abc}$ is $\frac{r+s+t}{2}$ while the power of $p$ in $\gcd(a,b,c)\cdot\mathrm{lcm}(a,b,c)$ is $r+t$, so that $r+t=s$. But $r\le s\le t$ then implies that $r=0$ and $s=t$. Therefore $\gcd(a,b,c)=1$. Further, this implies that if $p$ divides any of $a$, $b$, and $c$, then it divides exactly two of them, and to the same power.
Now, given $a$ and $b$ satisfying that condition (that is, that if a prime divides both $a$ and $b$, it divides them to the same power), it is easy to construct the only $c$ that works: take the product of the prime power factors unique to $a$ and $b$. Thus for example if $a = 8\cdot 27$ and $b = 27\cdot 25$, then we take $c=8\cdot 25$.
A: This is not a solution but maybe a way you can use to continue your analysis.
We assume a,b,c>0.
We have
$$\gcd(a,b,c)\cdot \text{lcm}(a,b,c)=\sqrt{abc}$$
and $a$,$b$,$c$ are products of prime powers so this must also hold for the prime powers. We have
$$\gcd(p^u,p^v,p^w)\text{lcm}(p^u,p^v,p^w)=\sqrt{p^u p^v p^w}$$
or
$$\min(u,v,w)+\max(u,v,w)=\frac{u+v+w}2$$
Without loss of generality we assume $u\le v \le w$ and we get
$$u=0, v=w$$
So for a prime $p$ and a power $e$ such that $p^e<30$  we have the triples $(a,b,c)=$
$$(1,p^e,p^e),(p^e,1,p^e),(p^e,p^e,1)$$ that satisfy the conditions. If we have two such triples $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ such that no prime divides $a_1a_2$,$b_1 b_2$ and $c_1c_2$ and that $a_1 a_2\le 30$,$b_1 b_2\le 30$,$c_1c_2\le 30$, then $(a_1 a_2,b_1 b_2,c_1c_2)$, is a solution, too.
So let's construct some solutions:
$2^2\le 30$, so $(2^2,1,2^2)$ is a solution
$5^1\le 30$, so $(5^1,5^1,1)$ is a solution
and also
$(2^2 5^1,5^1,2^2 )=(20,5,4)$
In a similar manner we find out that
$(2^1 5^1,3^1 5^1,2^1 3^1 )=(10,15,6)$
