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.
 A: (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 $k=2,\, m = \frac{n}{2}$.
A: Too long for a comment:
In addition to the rather lengthy
\begin{align}
&(m^2,\\
&((-1)^{2 k} \left(2 (-1)^k k m+(-1)^{k+1} (m+2)+m-6\right)^2)/16,\\
&\left((-1)^k \left(2 (-1)^k k m+(-1)^{k+1} (m+2)+m-6\right)+1\right)^2/4)\\
\end{align}
we also have $(a,b,c):$
\begin{align}
&\left(k^3+k^2+k+1,k^3+k^2+k+1,k^4\right)\\
&\left(k^4+k^2+1,k^4+k^2+1,k^6\right)\\
&\left(k m^2,k m^2 \left(k m^2-2\right)^2,\left(k m^2 \left(k m^2-3\right)+1\right)^2\right)\\
\end{align}
and for $f(n)=(n-1)^2$ we also have
\begin{align}
&\left(k^2,f^{2 n-1} \left((k m+1)^2\right),f^{2 n} \left((k m+1)^2\right)\right)\\
\end{align}
where $f^n$ is $f$ iterated $n$ times for $n \geq 1.$
However, even for fixed $a,$ the above formulae don't catch all of the solutions (and they say nothing of non-square $a$ combinations), and yet for each $a$ there seem to be multiple (infinite?) solutions.
Examples: case $a=8:$
A straightforward brute-force search for $(8,b,c);\ (b,c)<1000$ gives triples 
$(8,2,49),(8,8,49),(8,18,49),(8,18,289),(8,32,49),(8,32,961),(8,49,72),(8,49,288),(8,289,392),(8,392,529),$
where it is immediately apparent that the same numbers recur a number of times. Removing the $8$ and graphing shows the connectedness more clearly:

Searching for $c$ only, using the distinct elements from the initial search (eg $(8,49,c)$, etc.) up to $10^5$ reveals further connections: 

$(8,49,c)$ for example turns up $6$ triplets: $(8,49,2),(8,49,8),(8,49,18),(8,49,32),(8,49,72),(8,49,288)$ 
It may be more pertinent to ask then, are there infinitely many triplets for fixed $a?$ Certainly where $a$ is square, this is the case, but it is less clear whether this is the case when it is not.
It may also be worthwhile pursuing the idea of primitive pairs $(a,b).$
A: We study solutions, if any, of the shape $(a,b,c) = (x^2 P + 1,~ y^2 P + 1,~ z^2)$. We need 
$$z^2 \left(x^2 y^2 P^2 + (x^2 + y^2) P + 1 ) \right)$$
be a square. Namely, we need
\begin{align} 
2 x^2 y^2 P &= -(x^2 + y^2) + \sqrt{ (x^2 + y^2)^2 + 4(w^2 - 1)x^2 y^2} \\
\implies P &= \frac{ -(x^2 + y^2) + u }{ 2 x^2 y^2}
\end{align} 
where we have
$$(x^2 + y^2)^2 + (2wxy)^2 = (2xy)^2 + u^2$$
The equation $X^2 + Y^2 = V^2 + W^2$ like the Pythagorean one has also infinitely many solutions. For arbitrary integer $M,N,r,s$ we have
\begin{align}
2X &= r M + s N & 2Y &= r N – s M \\
2V &= r N + s M & 2W &= r M – s N
\end{align}
making 
\begin{align}
2(x^2 + y^2) &= r M + s N & 4wxy &= r N – s M \\  
4xy &= rN + sM & 2u &= rM – sN
\end{align}
The main equation 
$$(z^2 - 1) x^2 y^2 P^2 = Zzw \quad (?)$$ 
with $Z$ integer becomes 
$$4(z^2 - 1)s^2 N^2 = Zz (r^2 N^2 - s^2 M^2)$$
and we have to choose conveniently, if possible, among the arbitrary parameters in order this equation be verified, in particular $z$ must divide $4s^2 N^2$ for instance. No time for me now. If someone wants to continue go on. 
