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