Finding triplets $(a,b,c)$ such that $\sqrt{abc}\in\mathbb N$ divides $(a-1)(b-1)(c-1)$ When I was playing with numbers, I found that there are many triplets of three positive integers $(a,b,c)$ such that 


*

*$\color{red}{2\le} a\le b\le c$

*$\sqrt{abc}\in\mathbb N$

*$\sqrt{abc}$ divides $(a-1)(b-1)(c-1)$


Examples : The followings are positive integers.
$$\frac{(2-1)(8-1)(49-1)}{\sqrt{2\cdot 8\cdot 49}},\ \frac{(6-1)(24-1)(529-1)}{\sqrt{6\cdot 24\cdot 529}},\frac{(7-1)(63-1)(3844-1)}{\sqrt{7\cdot 63\cdot 3844}}$$
Then, I began to try to find every such triplet. Then, I found 
$$(a,b,c)=(k,km^2,(km^2-1)^2)$$
where $k,m$ are positive integers such that $k\ge 2$ and $km^2\ge 3$, so I knew that there are infinitely many such triplets. However, I can neither find the other triplets nor prove that there are no other triplets. So, here is my question. 

Question : How can we find every such triplet $(a,b,c)$?

Added : There are other triplets : $(a,b,c)=(k,k,(k-1)^4)\ (k\ge 3)$ by a user user84413, $(6,24,25),(15,15,16)$ by a user Théophile. Also, from the first example by Théophile, I got $(2k,8k,(2k-1)^2)\ (k\ge 3)$.
Added : $(a,b,c)=(k^2,(k+1)^2,(k+2)^2)\ (k\ge 2)$ found by a user coffeemath. From this example, I got $(k^2,(k+1)^2,(k-1)^2(k+2)^2)\ (k\ge 2)$.
Added : I got $(a,b,c)=(2(2k-1),32(2k-1),(4k-3)^2)\ (k\ge 5)$.
Added : I got $(a,b,c)=(k,(k-1)^2,k(k-2)^2)\ (k\ge 4)$.
Added : A squarefree triplet $(6,10,15)$ and $(4,k^2,(k+1)^2)\ (k\ge 2)$ found by a user martin. 
Added : user52733 shows that $(6,10,15)$ is the only squarefree solution.

(Too long for a comment.)
The two solutions,
$$a,\,b,\,c = k^2,\;(k+1)^2,\;(k+2)^2$$
$$a,\,b,\,c = 2^2,\;k^2,\;(k+1)^2$$
by users coffeemath and martin, respectively, are special cases of the more general solution,
$$a,\,b,\,c = k^2,\;(km\pm1)^2,\;(km\pm2)^2$$
where coffeemath's had $m=1$, while martin's had $k=2,\, m = \frac{n}{2}$.
