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

[all at abouthugo.de]

Here is a copy of an answer by Dave Rusin posted
in the parallel thread on sci.math. The start
of the thread was invisible in Google Groups and
Usenet. The Math Forum link is:
http://mathforum.org/discuss/sci.math/t/521337

Hugo

Re: a/b+b/c+c/a=n
Datum: 
14 Jul 2003 17:50:23 GMT
Von: rusin at vesuvius.math.niu.edu (Dave Rusin)
Northern Illinois Univ., Dept. of Mathematical Sciences
Foren: sci.math

In article <3F11C053.A83DB1A6 at abouthugo.de>,
Hugo Pfoertner  <nothing at abouthugo.de> wrote:

>>On 10 Jul 2003, Marcus wrote:
>>> I'm stuck on this problem that was posed on another
>>> message board.  The problem is to find which positive
>>> integers n can be the result of a/b + b/c + c/a, where
>>> a, b, and c are positive integers.

>>Anyway, an approach is to assume gcd(a,b,c)=1, though a, b,
>>and c may not be pairwise relatively prime.  Then it turns
>>out that we can find integers t,u,v that are pairwise
>>relatively prime, and for which a=t^2 u, b=u^2 v, c=v^2 t.

>>Anyway, n=(t^3+u^3+v^3)/(tuv).

Perhaps I'm simply retreading ground covered in another 
conversation about this problem but...

That equation describes (for each  n)  an elliptic curve. 
You want a (positive) integer solution on the curve. 
Equivalently, you can set, say, v=1  and look for rational solutions to
    x^3+y^3+1 = n xy
and then scale back to integers.

You can write the elliptic curve in canonical form if you like;
a transformation which works is 
   x = -1/2*(X*n^2-2*Y)/n/(3*X-27*n+n^4)
   y = -1/2*(X*n^2+2*Y)/n/(3*X-27*n+n^4)
giving the form
   Y^2 = X^3 + 1/4*n*(n^3-108)*X^2 - 9*n^2*(n-3)*(n^2+3*n+9)*X
                 -n^3*(n^2+3*n+9)^2*(n-3)^2
We can then check for the existence of rational points. There is
always torsion of order 3 corresponding to solutions with
{t,u,v}={1,-1,0}.

The curve is singular when n=3.
It has rank 0 but additional torsion when n=5.
It has rank 0 (no nontrivial rational points) when n=1,2,4,7,8,11,12,...
It has rank 1 (infinitely many nontrivial solutions) when
n=6,9,10,13,14, ...
These are in agreement with the table posted, which shows solutions:

>>      n   t   u     v
>>      3   1   1     1
>>      5   1   1     2
>>      6   1   2     3
>>      9   2   3     7
>>     10   5   7    18
>>     13   9  13    38
>>     14   2   7    13
>>     17   5  18    37

This skips over are the cases n=15 and n=16, which also have rank 1. 
For example, for n=15, we have the generator  (X,Y)=(11160, 1674000), 
which gives x = 1/3, y = -7/3, z = 1 (and then t=1,u=-7,v=3).
You wanted _positive_ t,u,v, which means we need to find a multiple
of the generator [*] which makes  x>0 and y>0. I believe there
are no such points. (All multiples of a point on the unbounded portion
of an elliptic curve in normal form will also be in the unbounded
portion.)
The same thing happens for n=16,20,31,35,36,40,44,...

The case n=41 is the first with rank 2, which doesn't mean very much
in this setting except that there will be "more than the usual"
number of solutions (t,u,v). (The next rank-2 case is n=69.)

The behaviour for all the values of n < 67 mentioned in the previous
post
follows the pattern of one of the preceding paragraphs, except that
I was unable to verify that there were no solutions when n=62 and n=64
(I suppose the ranks are one but my software found no generator for the 
curves in the little time I allotted, so I couldn't check to see whether
the generator lay on the bounded part of the curve or not.)

dave

[*] Torsion doesn't help; adding one of the elements of order 3 just
permutes t,u,v cyclically. And taking negatives on the elliptic curve
corresponds to performing a permutation of odd sign on these three.