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

Olivier Gerard ogerard at ext.jussieu.fr
Sat Jul 12 23:09:46 CEST 2003

```Le 12, Hans Havermann écrivait:
> On Friday, July 11, 2003, at 02:58  PM, <all at abouthugo.de> wrote:
>
> >I had set the threshold for accepting a sum being equal to the nearest
> >integer to 10^(-9)
> >and the 3 wrong solutions (=4) all had absolute differences < 10^-9.
> >results?

The problem is not high precision but factorization. If you stay within
reasonable bounds for your experiments, you should not try to compute the
individual fractions and sum them but compute the denominator and the numerator of
the sum of the three, which are respectively
a b c  and   a^2 c + b^2 a + c^2 b
and factor them (or let your software reduce the fraction for you).

Here are the integers representable for distinct positive a,b,c

with (a*b*c)^(1/3)  <= 1000
5,6,9,10,14,19,21,41,53,66,149,269

with (a*b*c)^(1/3)  <= 10000
5,6,9,10,13,14,17,19,21,30,41,51,53,54,66,69,83,106,149,154,174,178,237,
250,261,269,633,978,1769,2309,2369,14803,21529

Just add 3 to these lists if you allow a,b and c to be equal.

> I think your (=7) results are wrong as well. Can it be shown that, as a
> minimum, all solutions require (a*b*c) to be a cube?
>
Yes and in fact much more is true. See the recent mail by Dean
for a proof of this, and a reference to the OEIS.

Now, let's consider the problem from the point of view of
which cubes (a*b*c) can be decomposed into a triple (a,b,c).

Let's call a triple (a,b,c) primitive if
a, b and c all different, positive and GCD(a,b,c) = 1.

Here is the list of the first r = (a*b*c)^(1/3) for which there exists
such primitive triples

(2) 6 18 42 45 70 182 378 546 630 854 1206 1302 1314 1330 1554 1638 1890 ...
(not in the OEIS)

and here is the corresponding number of different primitive triples

(1) 2 2 4 2 2 2 2 2 4 2 2 2 2 2 2 4 2 2 2 2 2 2 2 ...
(not in the OEIS)

and the number of different values represented by these triples
is the same sequence divided by 2 except for the first term

The smallest odd example
is for  a*b*c = 45^3,  n=19 for (5,225,81) and  (9,405,25)

The first case for more than one integer represented is 42 giving
9 and 66 through 2x2 non equivalent triples.

**  Let (a,b,c) be a primitive triple
If we call  (i,j,k)= (GCD(a,b), GCD(b,c), GCD(c,a))
and r= (a*b*c)^(1/3)  then   r= i*j*k

Furthermore GCD(i,j,k)=1.

(Note: there are other conditions necessary for (i,j,k))

Here is a sample list grouped by 2 under the
transposition (i,j,k) -> (i,k,j).

r	n	i j k		a b c
=========================================================
2	5	1 2 1		1 4 2
6	6	1 2 3	1 3 2	3 4 18  	2 9 12
18	41	1 2 9	1 9 2	81 2 36 	9 4 162
42	9	2 3 7	2 7 3	28 18 147	12 98 63
42	66	1 3 14	1 14 3	14 9 588	3 196 126
45	19	1 5 9	1 9 5	9 25 405	5 81 225

etc...

We note that in the case of 6 and 45, the triples (a,b,c)
are in ascending order when the corresponding (i,j,k)
is aligned on the least member, and that it is not the
case for the others listed here.

Olivier

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