# 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)

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.

Then, I began to try to find every such triplet. Then, I found

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:

In addition to the rather lengthy

we also have $(a,b,c):$

and for $f(n)=(n-1)^2$ we also have

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 $(a,b).$