From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: a + b + c = 1 = a*b*c Date: 7 Feb 2000 16:02:44 GMT Newsgroups: sci.math Summary: [missing] The subject line has been generalized in other posts to (what is equivalent to) a+b+c=1 and a*b*c=n. In this case we modify this approach: In article <21197-389B8A55-27@storefull-126.bryant.webtv.net>, Oscar Lanzi III wrote: >You have a+b+c = 1 and a*b*c = 1. Rearrange these equations to give: > >a+b = 1-c >a*b = 1/c a*b = n/c From the root-coefficient equations for polynomial equations. you may >conclude that a and b are the roots of > >x^2 - (1-c)*x + (1/c) = 0. x^2 - (1-c)*x + (n/c) = 0 >These roots will be rational if and only if the discriminant is a >rational square. So now you have the single rational-variable >diophantine equation > >(1-c)^2 - 4/c = d^2 (c, d rational) (1-c)^2 - 4*n/c = d^2, or equivalently the 2-variable cubic c^3 - c*d^2 - 2*c^2 + c - 4*n = 0. Or you might replace c by -4*n/x and d with y/x and get the equation y^2 = x^3 + x^2 + (8*n)*x + (16*n^2) >Now does THIS have a solution? I don't know, but I'm not optimistic.. Depends on n, of course. For each n (other than n=0 and n=1/27) this describes an elliptic curve. For all n there are two points with x=0 ; these describe a torsion subgroup of order 3, but these points don't correspond to any values of c and d. So the question is whether there are other rational points besides this torsion subgroup. For n=1, 2, 4, 6, 7, ... the answer is no. For n=3 there are additional torsion points with x=12 and -4; that's it. The case n=9 is similar. For n=5, the curve has rank 1; a generator is the point with x=-5. Likewise when n=8 we have rank=1 and a generator with x=-8, and when n=10 we have a generator with x=15. (I didn't find any examples with rank 2 but then, I only checked for integer values of n between -10 and 10.) dave ============================================================================== From: "ERIK DOFS" Subject: Re: a + b + c = 1 = a*b*c Date: Sat, 5 Feb 2000 13:16:02 +0100 Newsgroups: sci.math This problem was first solved by Cassels in 1960 ('On a diophantine equation', Acta Arithmetica 6,1960-61), and the equation has no non-trivial solutions. The general Diophantine problem x^3+y^3+z^3 = nxyz where n is any integer has been treated by many mathematicians, since JJ Sylvester (the Fermat case n=0 was treated by Euler). I have solved the equation for several infinite classes of n, and a table of non-solvable cases can be found in 'Unsolvable cases of P^3+Q^3+cZ^3=dPQR' in the Rocky Mountain Journal Of Mathematics', Vol.28, Nr 2, summer 1998. Ray skrev i meddelandet <389B5BA5.FC9071B@stupid.com>... >Are there any rational solutions to a + b + c = 1 = a*b*c? >We can write this as k + l + m = d, k*l*m = d^3, where k,l,m,d are >integers. >Then, we can assume that k,l,m share no common factor. If k and d share >a common factor, say p, then p can't divide l, since if it did, it >couldn't divide m, but then p|d-k-l, contradiction. >So k,l, and m are pairwise relatively prime. >But then k,l, and m are all cubes. So we can write this as: >x^3 + y^3 + z^3 = x*y*z = d. >Now, either one or all of x,y,z must be divisible by 3, but I'm not sure >if that helps. And of course, one must be positive, the other two >negative. >Anyway, that's as far as I got. >And no, this isn't homework. ============================================================================== From: Pierre Bornsztein Subject: Re: a + b + c = 1 = a*b*c Date: Tue, 08 Feb 2000 00:04:08 +0100 Newsgroups: sci.math ERIK DOFS wrote: > > This problem was first solved by Cassels in 1960 ('On a diophantine > equation', Acta Arithmetica 6,1960-61), and the equation has no non-trivial > solutions. The general Diophantine problem x^3+y^3+z^3 = nxyz where n is any > integer has been treated by many mathematicians, since JJ Sylvester (the > Fermat case n=0 was treated by Euler). I have solved the equation for > several infinite classes of n, and a table of non-solvable cases can be > found in 'Unsolvable cases of P^3+Q^3+cZ^3=dPQR' in the Rocky Mountain > Journal Of Mathematics', Vol.28, Nr 2, summer 1998. Some complements : Cassels' proof has been simplified by Sansone and Cassels see : "Sur le probl�me de M.Werner Mnich", Acta arithmetica (1962, p.187-190). Cassels proof uses infinite descent in Z[j]. Some generalizations : 1) A.Schinzel has proved : For each integer s > 3, the equation x_1 + x_2 + ... + x_s = x_1 * x_2 * ... * x_s = 1 has an infinite number of solutions in rational numbers. see : Elemente der Mathematik (1956, p.134). 2) On the diophantine equation x^3 + y^3 + z^3 + kxyz = 0 see : - Mordell "Diophantine equation" Academic Press. - Barry Mazur in Inventiones Math 18 (1972) p.183-266 Pierre. > >Are there any rational solutions to a + b + c = 1 = a*b*c? > >We can write this as k + l + m = d, k*l*m = d^3, where k,l,m,d are > >integers. ... So we can write this as: > >x^3 + y^3 + z^3 = x*y*z = d. ============================================================================== From: "ERIK DOFS" Subject: Re: a + b + c = 1 = a*b*c Date: Wed, 9 Feb 2000 20:42:41 +0100 Newsgroups: sci.math Ray skrev i meddelandet <389FBFC4.C7501B1D@stupid.com>... >I wrote: >> >> Are there any rational solutions to a + b + c = 1 = a*b*c? > >Thanks to all who replied, especially ERIK and Pierre for those >references. If somebody is interested in SOLVABLE cases of x^3+y^3+z^3=nxyz, I recommend (my own): 'Solution of x^3+y^3+z^3=nxyz' in Acta Aritmetica 73 (1995). E.g., there are parametric solutions for n=-m^2, m^2+5, -(m^2+15)/4 (m odd), -(m^2+24), (m^2+33)/2 (m odd), (m^2+255)/4 (m odd), -(m^2+8), (m^4+15m^2+48)/4,....as well as gigantic solutions for small n: n=73, x,y,z=89200900157319, 2848691279889518, 1231526622949983 (if I remember correctly...).