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

[Dean Hickerson]

Jud McCranie pointed out that I messed up again!

Hans Havermann asked:

> Is that second triple correct?

I said:

> Oops, no it's not.  It should be  (e f^2, f d^2, d e^2).

Right numbers, wrong order.  It should be  (e f^2, d e^2, f d^2).

To convince myself that I've got it right this time, let's look at an
example:  For  (d,e,f) = (2,3,7),  we have  n = (d^3+e^3+f^3)/(d e f) = 9.
This leads to the two triples

    (a,b,c) = (e^2 f, f^2 d, d^2 e) = (63, 98, 12)

and

    (a,b,c) = (e f^2, d e^2, f d^2) = (147, 18, 28);

note that

    63/98 + 98/12 + 12/63 = 147/18 + 18/28 + 28/147 = 9,

as expected.

My program has reached max(d,e,f)=19741; I've updated the list at

    http://www.math.ucdavis.edu/~dean/defn.txt

This includes a new solution for n=6 besides (d,e,f) = (1,2,3):
(d,e,f) = (1817, 3258, 5275).

Dean Hickerson
dean at math.ucdavis.edu