a/b + b/c + c/a = n

[Dean Hickerson]

Hans Havermann asked:

> Can it be shown that, as a minimum, all solutions require (a*b*c) to be
> a cube?

Yes:  Without loss of generality,  gcd(a,b,c)=1.  Let  d=gcd(b,c),
e=gcd(c,a),  and  f=gcd(a,b).  Then d, e, and f are relatively prime in
pairs.  Also, a is divisible by e and f, so it's divisible by ef; let
a = e f A.  Similarly, let  b = f d B  and  c = d e C.

We have  d = gcd(b,c) = gcd(f d B, d e C) = d gcd(f B, e C), so B is
relatively prime to both e and C.  Similarly, C is r.p. to both f and A,
and A is r.p. to both d and B.  In other words, A, B, and C are r.p. in
pairs, and  gcd(A,d) = gcd(B,e) = gcd(C,f) = 1.

Also,

                         2    2      2    2      2    2
                      A B  d f  + B C  e d  + C A  f e
    a/b + b/c + c/a = ---------------------------------
                                 A B C d e f

So A divides  B C^2 e d^2,  but is relatively prime to B, C, and d, so
A divides e.  Also, e divides  A B^2 d f^2,  but is r.p. to d, f, and B,
so e divides A.  Hence A=e.  Similarly, B=f and C=d.  Hence:

    a = e^2 f,   b = f^2 d,  c = d^2 e,

so  a b c = (d e f)^3.  Also,

                       3    3    3
                      d  + e  + f
    a/b + b/c + c/a = ------------.
                         d e f

Now we can look for integer values of  a/b + b/c + c/a  more efficiently:
Look for triples (d,e,f) with  1 <= d <= e <= f,  relatively prime in pairs,
such that  d^3 + e^3 + f^3  is divisible by  d e f.  For each such triple, we
get two triples (a,b,c):  (e^2 f, f^2 d, d^2 e)  and  (e f^2, d f^2, e d^2).

Incidentally, the first few values of n are given by sequence A072716:

%S A072716 3,5,6,9,10,13,14,17,18,19,21
%N A072716 Integers expressible as (x^3 + y^3 + z^3)/xyz with positive
           integers x, y, and z. (Alternatively, the integers expressible
           as x/y + y/z + z/x with positive integers x, y, and z.)
...
%A A072716 Tadaaki Ohno (t-ohno(AT)hyper.ocn.ne.jp), Aug 07 2002

Apparently the author doesn't know if 22 is in the sequence.

Dean Hickerson
dean at math.ucdavis.edu