Finding triplets (a,b,c)(a,b,c) such that √abc∈N\sqrt{abc}\in\mathbb N divides (a−1)(b−1)(c−1)(a-1)(b-1)(c-1)

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 $(a,b).$