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

Mon Jul 14 22:48:02 CEST 2003

```Here is a copy of an answer by Dave Rusin posted
in the parallel thread on sci.math. The start
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

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

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.

```