We investigate the solutions of X^3+Y^3+Z^3 = 0. First note that if F = u^2*v+v^2*w+w^2*u G = u*v^2+v*w^2+w*u^2 H = u*v*w (or any fixed multiples thereof) then F^3 + G^3 + 9*H^3 - 6*F*G*H = (u^3+v^3+w^3)*(u^3*v^3+u^3*w^3+v^3*w^3) Similarly if X = (F^3-G^3+9*H^3)/2 Y = (G^3-F^3+9*H^3)/2 Z = (F^3+G^3-9*H^3)/2 then X^3+Y^3+Z^3 = 1/8*(F^3-6*F*G*H+G^3+9*H^3)* (F^6+6*F^4*G*H+2*F^3*G^3+18*F^3*H^3+36*F^2*G^2*H^2+ 6*F*G^4*H+54*F*G*H^4+G^6+18*G^3*H^3+81*H^6) Thus u^3+v^3+w^3=0 => F^3+G^3+9*H^3=6*F*G*H and F^3+G^3+9*H^3=6*F*G*H => X^3 + Y^3 + Z^3 = 0. We therefore view these formulas as describing maps between the projective varieties V = { [x,y,z] : x^3+y^3+z^3 = 0 } and W = { [x,y,z] : x^3+y^3+9z^3 = 6xyz } (Reduced to rational affine varieties and then put in canonical form, these are respectively the elliptic curves B27: y^2=x^3-432 and D27 in Cremona's list; see APECS). The composite of these maps sends [u,v,w] to [v^9-3*v^6*w^3-6*v^3*w^6-w^9, -v^9-6*v^6*w^3-3*v^3*w^6+w^9, -3*(u^6-v^3*w^3)*u*v*w], which happens to be the triple of [u,v,w] in the elliptic curve. (Actually, that's backwards. I transformed the curve to a rational affine curve, put it in WNF, explicitly computed the triple, and transformed back. I suspected that the transformations FGH->XYZ, which I knew were surjective, factored through the tripling map, so I used this to compute FGH in terms of uvw under that assumption.) We conclude that if V has any rational points besides [0,1,-1] [-1,0,1] and [1,-1,0], then W also has rational points and hence (barring 9-torsion) V actually has positive rank. The real work goes the other way. Elementary considerations show that if [X,Y,Z] is a point on V in lowest terms, then there exist integers FGH with X+Z=F^3, Y+Z=G^3, X+Y=9H^3 (assuming Z is the one divisible by 3 -- one must exist); that is, the map W -> V is onto over Q. So W (torsion={1}) must also be shown to have no real points BUT NOW it is more or less sufficient to show uvw -> FGH is onto... I guess I need to define uvw : u = gcd(H, ...)? show H^3=(uvw)^3, =-(1/9)(F^3+G^3-6FGH) factors into 3 rel prime parts, thus each a cube, call them u^3 v^3 w^3... ?? F^2+FG+G^2 = (v^2+vw+w^2)()(), assuming uvw->FGH onto... So I guess -(F^2_FG_G^2)=(v^2-vw+w^2)()(). Hm; multiply by (v+w)()() to get (v^3+w^3)()() = (uvw)^3 = H^3 ( (u+v+w)^3=3(u+v)(u+w)(v+w) if u^3+v^3+w^3=0... ) Note a problem: [u,v,w] are not unique; cyclic permutations work as well... (That is, there is a kernel of order 3 in V->W, namely the rational torsionpts. We can try for the _elem.sym.polys_ of {u,v,w}; if they're STU we have S = ( F+G + 3H)/T U = H but T seems to be a cubic; taking s=F+G, t=(F-G)^2, u=H we need 3 2 2 3 6 4 u (s + 3 u) + (t - s - 24 s u - 36 u ) T + 4 T = 0 i.e. T is a root of 4*u*(s+3*u)^3+(t-s^2-24*s*u-36*u^2)*T^3+4*T^6 . with F=u^2v etc, G, H we did > {F/H-f,G/H-g,S-(u+v+w),T-(u*v+v*w+w*u),u^3+v^3+w^3}; > subs(w=-1,"); > eliminate(",{u,v}); The relation is rel:= f^3 + g^3 -6*g*f + 9 ; So here "S" = u+v+w = u+v-1; "T" = u*v +w*(u+v) = u*v - (u+v); so what we really want is Try subs({f=inps[1]/inps[3],g=inps[2]/inps[3],S=inps[4],T=inps[5]},"); where inps:=[-.175710029659194241659628722587, 1.16387880513906599025431758354, -.478232795693097275299779209888, .45646559138619455059955841978, -.978232795693097275299779209887, -.478232795693097275299779209888, 0] (F,G,H,S,T,uvw,u^3+v^3+w^3); u=1/2, w=-1, v=(7/8)^(1/3) T = 1/3*(S^3-3*S-3)/(S+1) seems consistent, leads to [2 eqns] Really the right eqn follows from factoring the diff of the two elements: ah:= 3*f+6+3*g+(3*f+6+3*g)*S+(f+3+g)*S^2+S^3 ; since both parts of above are in ideal gen'd by this and rel. ============================================================================== the formulas u = (1+Y)/(2*X) v = (1-Y)/(2*X) and X = 1/(u+v) Y = (u-v)/(u+v) exhibit a one-to-one correspondence between the solutions {u,v} of u^3 + v^3 = 1 and those of 3Y^2 = 4X^3 - 1 not having X = 0. This calculation is valid over any field of characteristic not 2, and so applies in particular to the rational numbers (where the equation 3Y^2 = -1 has no solution). This is relevant for Fermat's Last Theorem. On the one hand, for any integer solution of x^3 + y^3 = z^3 not having z=0 we may divide by z^3 and obtain a rational solution to u^3 + v^3 = 1, and conversely we can clear denominators in any rational solution of u^3+v^3=1 to get an integer solution to x^3+y^3=z^3. (This reasoning applies to any homogeneous Diophantine equation(s): there is no real difference between integral and rational points on a projective variety.) Very well, then, this SINGLE substitution now shows that if there were a nontrivial integral solution to the Fermat equation with p=3, there would be a nontrivial rational solution to 3Y^2 = 4X^3 - 1. (Trivial integral solutions are those with xyz=0 and correspond to the rational solutions with X=1 and Y = +- 1. =============================================================================