From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Knot of congruences - revisited
Date: 15 Apr 1995 06:26:34 GMT

Wouldn't you know it, the day I get around to the problem it rolls off
our newsreader. Anyway, there was a post asking for solutions (or proof
of lack thereof) to the congruences
	ab=c mod (a+b)
	bc=a mod (b+c)
	ca=b mod (c+a)

This is an unusual problem, but I can make it more unusual yet!
Here's an analysis which leads to no definitive solution, but may 
point in the right direction. (Since I don't have a strong conclusion
to draw, I didn't double-check the algebra carefully, but the ideas
are sound.)

Let's see, we seek pairwise relatively prime positive integers a,b,c
making each of these rational values integral:

	x=(ab-c)/(a+b), y=(bc-a)/(b+c), z=(ca-b)/(c+b)

Well, given _any_ complex numbers  a, b, and c, I can compute these
three numbers  x, y, and z: they are the roots of the polynomial
X^3 - S1 X^2 + S2 X - S3, where the  Si  are the elementary symmetric
polynomials in  x, y, and  z.  It's clear before we begin that we can
compute these in terms of the corresponding polynomials  s1, s2, and s3
in  a, b, and c  (s1=a+b+c, etc.) Here's the result of the calculation:

	S1= 3-s1+ (s2-s1)(s1^2+s2)/(s1s2-s3)

	S2= 3+s1-2s2 + s2 (s2-s1)(2)(s1+1)/(s1s2-s3)

	S3= (1+s1)^2-(1+s1)s2-s3 + s2 (s2-s1)(1+s1)^2/(s1s2-s3)

where the denominators all equal (a+b)(a+c)(b+c).  Of course, one can likewise
compute  a, b, and  c  as roots of a polynomial, so the problem may be
recast as follows:  Find complex numbers  s1, s2, s3  so that the roots
of both
    X^3 - s1 X^2 + s2 X - s3    and    X^3 - S1 X^2 + S2 X - S3
are integers, the roots of the first polynomial positive and coprime.

Well, that's being silly: I needn't mention "complex numbers" since clearly
only integral  s1, s2, s3  will do. Also, in the presence of the assumption
that the roots of the first are integral, the statement that they be
pairwise coprime is equivalent to  gcd(s2,s3)=1. Finally, the positivity
of the roots is equivalent to a collection of inequalities on the  s_i
(s1>0, s2 > 0, etc.)

So really, we seek positive integers  s1, s2, s3 (subject to some 
inequalities and with gcd(s2,s3)=1) which make the roots of the two
polynomials rational integers.

Clearly, the second polynomial can only have integral solutions if
it has integral coefficients. Note that the gcd's  (s2,s1s2-s3)=
(s2,s3)=1, so that integrality of the coefficients is equivalent to
that of the final fractions listed in  S1 S2 and S3.

By calculating a few gcd's and so on, I deduce that this holds only
for integers of the following forms:
	s1= -1 + k z
	s2= -1 + k z + ij z
	s3= s1 s2 - i z^2
where  i,j,k, and  z  are integers. (Actually, the  "2" in the numerator
of  S2  also allows  k  to be a half-integer if  z  is even.)
Here the  gcd  (s2, s3) = (i, k z -1 )  must be  1. These  s_i  can only
lead to positive integer roots if  i,j,k, and  z  are positive.
There is a uniqueness to such a condition if we assume (j,z)=1.
For example, when (a,b,c)=(5,7,11), (s1,s2,s3)=(23,167,385), and
(i,j,k,z)=(6,1,1,24)

Thus the original problem may be recast as follows: 

| Find positive (half-) integers  i, j, k, z  (subject to some inequalities 
| and a gcd condition) so that the two integral polynomials
|	X^3 - (-1+kz) X^2 + ...       and
|	X^3 - (4-kz-jk+ij^2+jk^2z) X^2 + ...
| have rational roots.

There is no need to specify "integral roots" since the roots will now
automatically be algebraic integers.  Thus I have transformed the problem:
instead of seeking sets of integers making some rational numbers
integral, we need to find sets of integers making some integral
numbers rational!

Possibly this description of the problem can lead more rapidly to any
other examples of solutions -- it may be that you are searching a
more populated sample space when looping over all the (i,j,k,z) than
when looping over all the (a,b,c).

Then again, maybe not.

dave