## Complete cubic parametrization of the Fermat cubic surface w3 + x3 + y3 + z3 = 0

This is a famous Diophantine problem, to which Dickson's History of the Theory of Numbers, Vol. II devotes many pages. It is usually phrased as w3+x3+y3=z3 or w3+x3=y3+z3, with the implication that the variables are to be positive, as in the integer solutions 33+43+53=63 (an amusing counterpart of the classic identity 32+42=52 -- but no, it does not generalize further) and 123+13=103+93=1729 (which famously occurs in the Ramanujan taxicab story). If we allow negative variables, the three forms of the equation become equivalent by sign changes, and we choose the form w3+x3+y3+z3=0 that displays the greatest symmetry. The two solutions above are then (w,x,y,z)=(3,4,5,-6) and (x,y,z,t)=(12,1,-10,-9). These, together with (1,6,8,-9) and variations obtained by scaling or permuting the variables, are minimal when we exclude trivial solutions of the form (x,-x,y,-y) and permutations thereof.

Theorem. Suppose (w,x,y,z) is a rational solution of w3+x3+y3+z3=0. Then there exist r,s,t such that (w,x,y,z) are proportional to

-(s+r)t2 + (s2+2r2) t - s3 + rs2 - 2r2s - r3,
t3 - (s+r)t2 + (s2+2r2) t + rs2 - 2r2s + r3,
-t3 + (s+r)t2 - (s2+2r2) t + 2rs2 - r2s + 2r3,
(s-2r)t2 + (r2-s2) t + s3 - rs2 + 2r2s - 2r3.
Indeed, r,s,t are proportional to
yz - wx,       wy - wx + xz + w2 - wz + z2,       y2 - xy + wy + x2 - wx + xz,
unless w+y, x+z are both zero, in which case r,s,t are proportional to x+y,y,x.

This map (w:x:y:z) -> (r:s:t) blows down six disjoint lines, as follows:

 w+az = x+ay = 0 to (-a:1:1); w+a'z = x+a'y = 0 to (-a':1:1); x+az = y+aw = 0 to (0:1:-a); x+a'z = y+a'w = 0 to (0:1:-a'); y+az = w+ax = 0 to (1:-a':-a); y+a'z = w+a'x = 0 to (1:-a:-a');
where a,a' are a cube root of unity and its complex conjugate. We have thus explicitly represented the Fermat cubic surface as P2 blown up at six points; I do not know if such formulas have been previously exhibited. The usual approach to rationally parametrizing that surface, using a Q(a)-conjugate pair of skew lines, amounts to getting that surface as (P1)2 blown up at 5 points.

Of the fifteen lines connecting six of the blown-up points, three are defined over Q. We have already seen (s+t:s:t) mapping to (s:-t:-s:t); the other two lines are (0:s:t), which maps to (s:-t:t:-s), and (r:s:s), which maps to (-x:x:y:-y) with x=-(r+s), y=2r-s.

While we stated the result only for rational solutions, it holds equally over any field F not of characteristic 3, as long as F does not contain the cube roots of unity a,a'. If F does contain them, we must exclude points (w,x,y,z) on one of the six blown-down lines, and also give separate formulas for r,s,t on two further lines.

The surface w3+x3+y3+z3=0 has 4! obvious symmetries, obtained by permuting the variables. It turns out that the even permutations preserve the set of six blown-down lines, and thus give rise to linear transformations of the (r:s:t) projective plane. This is not obvious from our formulas; the relevant symmetries of P2 can be seen more clearly after the linear change of variable

(A,B,C) = (r,s-t,-r+s+t);     (r,s,t) = (2A,A+B+C,A-B+C)/2 .

This makes our formulas longer (no coefficients vanish), and also ill-behaved in characteristic 2; but has the advantage that the symmetries of P2 become monomial permutation matrices. Thus: let F(A,B,C) be the homogeneous cubic

9A3 + 9A2B + 3A2C + 3AB2 - 6ABC + 3AC2 + 3B3 + 3B2C + BC2 + C3.

Then w,x,y,z are proportional to

F(A,B,C),    F(-A,B,-C),    F(-A,-B,C),    F(A,-B,-C).

This displays the four-group symmetry as diagonal 3*3 matrices of determinant 1. Moreover, from the pattern of coefficients of F:

9
9     3
3   -6     3
3     3     1     1
we see that F(A,31/3B,32/3C) is invariant under cyclic permutations of the variables. In other words, 3F(C,A,B)=F(A,B,3C). This yields cyclic permutations of x,y,z; together with the four-group, we obtain the full alternating group on {w,x,y,z}.

The odd permutations of {w,x,y,z} send the six lines to a different set of six disjoint lines on the cubic surface, namely the set that forms a ``double six'' with the first set.