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

  • 2abc
  • abcN
  • abc divides (a1)(b1)(c1)

Examples : The followings are positive integers.
(21)(81)(491)2849, (61)(241)(5291)624529,(71)(631)(38441)7633844

Then, I began to try to find every such triplet. Then, I found
(a,b,c)=(k,km2,(km21)2)
where k,m are positive integers such that k2 and km23, 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,(k1)4) (k3) 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,(2k1)2) (k3).

Added : (a,b,c)=(k2,(k+1)2,(k+2)2) (k2) found by a user coffeemath. From this example, I got (k2,(k+1)2,(k1)2(k+2)2) (k2).

Added : I got (a,b,c)=(2(2k1),32(2k1),(4k3)2) (k5).

Added : I got (a,b,c)=(k,(k1)2,k(k2)2) (k4).

Added : A squarefree triplet (6,10,15) and (4,k2,(k+1)2) (k2) found by a user martin.

Added : user52733 shows that (6,10,15) is the only squarefree solution.

Answer

Too long for a comment:

In addition to the rather lengthy

(m2,((1)2k(2(1)kkm+(1)k+1(m+2)+m6)2)/16,((1)k(2(1)kkm+(1)k+1(m+2)+m6)+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(km22)2,(km2(km23)+1)2)

and for f(n)=(n1)2 we also have

(k2,f2n1((km+1)2),f2n((km+1)2))

where fn is f iterated n times for n1.

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

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