When I was playing with numbers, I found that there are many triplets of three positive integers (a,b,c) such that
- √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≥2 and km2≥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≥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≥3).
Added : (a,b,c)=(k2,(k+1)2,(k+2)2) (k≥2) found by a user coffeemath. From this example, I got (k2,(k+1)2,(k−1)2(k+2)2) (k≥2).
Added : I got (a,b,c)=(2(2k−1),32(2k−1),(4k−3)2) (k≥5).
Added : I got (a,b,c)=(k,(k−1)2,k(k−2)2) (k≥4).
Added : A squarefree triplet (6,10,15) and (4,k2,(k+1)2) (k≥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 fn is f iterated n times for n≥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
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 105 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).