When I was playing with numbers, I found that there are many triplets of three positive integers (a,b,c) such that

- 2≤a≤b≤c
- √abc∈N
- √abc divides (a−1)(b−1)(c−1)

Examples: The followings are positive integers.

(2−1)(8−1)(49−1)√2⋅8⋅49, (6−1)(24−1)(529−1)√6⋅24⋅529,(7−1)(63−1)(3844−1)√7⋅63⋅3844Then, I began to try to find

everysuch triplet. Then, I found

(a,b,c)=(k,km2,(km2−1)2)

where k,m are positive integers such that k≥2 and km2≥3, so I knew that there areinfinitelymany 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 findeverysuch 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 onlysquarefreesolution.

**Answer**

*Too long for a comment:*

In addition to the rather lengthy

(m2,((−1)2k(2(−1)kkm+(−1)k+1(m+2)+m−6)2)/16,((−1)k(2(−1)kkm+(−1)k+1(m+2)+m−6)+1)2/4)

we also have (a,b,c):

(k3+k2+k+1,k3+k2+k+1,k4)(k4+k2+1,k4+k2+1,k6)(km2,km2(km2−2)2,(km2(km2−3)+1)2)

and for f(n)=(n−1)2 we also have

(k2,f2n−1((km+1)2),f2n((km+1)2))

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

(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 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).

**Attribution***Source : Link , Question Author : mathlove , Answer Author : martin*