The above representation problem will probably strike a chord with anybody who has taught algebra, where some students seem certain that
 

( x + y + z )3  =  x3 + y3 + z3

for all x,y,z.

Combining the title equation into a single fraction, we get

(N-1)x3- 3(y+z)x2 - 3(y+z)2 x + (y+z)((N-1)y2-(N+2)y z +(N-1)z2)=0

On first sight, this cubic seems to offer few opportunities for simplification. If, however, we define s=x+y+z, it reduces to

3 N (s-y) x2 - 3 N (s-y)2 x + s(N(s2-3s y+3y2)-s2) = 0

If we wish to find integer solutions to the original problem, this quadratic in x must have integer solutions, so the discriminant must be an integer square.

Thus there must exist integer solutions to the quartic

d2 = 9N2y4 - 18N2s2y2 + 12N(N-1)s3y - 3N(N-4)s4

and if we define e=d/y² and g=s/y, there must be rational solutions to the quartic

e2 = 3 N(4-N) g4 + 12N(N-1)g3 - 18N2g2 + 9N2

There is the obvious solution g=0,e=3N, so the quartic is birationally equivalent to an elliptic curve.

Using standard transformations, we find that the elliptic curve is

EN: v2 = u3 + 36N2(u+6N(N-1))2

with the transformation

g =  s y  =  6 N u v - 36N3 + 36N2

As an example, consider N=29, which gives the curve

v2 = u3 + 30276u2 + 295009344u + 718642761984

on which we find the point u = 5050444/625, v = 36999062128/15625. This gives s/y = 1237575/1338064, and after some simple scaling calculations we find

x = 865373775 , y = 424166288 , z = -897228788.

The discriminant of the curve is

-212   39   N8   (N-1)3   (N-9)

so that the curve is singular for N=0 (solution x=2,y=z=-1), N=1 (solution x=1,y=z=0), N=9 (solution x=y=z=1).

Elliptic curves all consist of an infinite component, and possibly a separate closed "egg". In this problem, this depends on the number of real roots of

f(x)=x3+36N2(x+6N(N-1))2=0

Standard analysis shows that there are 3 real roots only for
1 < N < 9, if we ignore the singular curves.
 

We assume from now on that E_N is non-singular and investigate first the torsion points, which form a finite subgroup. The point at infinity is assumed to be the identity of the group of rational points, so we look for finite torsion points.

Clearly there are 2 rational points at (0, ±36N²(N-1)), which can be shown to be points of inflexion of the curve and so are points of order 3.

The points of order 3 imply, from Mazur's theorem, that the torsion subgroup is restricted to be isomorphic to one of Z3, Z6, Z9, Z12, Z2 x Z6.

The last possibility can only occur if there are 3 integer roots of f(x). Solving f(x)=0 for 1 < N < 9, we find this never happens. In fact there is only one case where we have ever found one integer root, namely N=4 with x=-144, where the torsion subgroup has 6 elements. This gives the fairly obvious solution x=1,y=1,z=0.

Numerical testing leads to the following conjecture:

The torsion subgroup is isomorphic to Z3 for all N except 1,4,9.

Can any reader prove this?

If there are only these 3 obvious torsion points, then none of them lead to a solution of the original problem. We thus must look for curves with rank greater than zero.
The rank of the elliptic curve was estimated using the calculations involved in the Birch and Swinnerton-Dyer conjecture.The ranks were estimated for
2 <= N <= 199, and for those curves with rank greater than zero, attempts were made to determine the rational coordinates of a generator. From these it is easy to determine (x,y,z) solution.

The results found so far are given in the file cuberes.

The methods used to find points were

• If the height is small, simple searches.
• For larger heights, an implementation of Silverman's canonical height method.
• For much larger heights, use of Heegner points, but only for curves of reasonable conductor.

One method which might help is to use the fact that the points of order 3 induce a 3-isogeny between E_N and an elliptic curve of the same rank, which might have a generator with height one-third that if E_N.

Using the standard method of Velu, we just follow the algebra to get the 3-isogenous curve

V2 = U3 + 36N2U2 + 3888N3(1-N)U + 3888N4(1-N)(41N-9)

where

U =  u3 + 864N3(N-1)u + 5184N4(N-1)2 u2

V = v  u3 + 864N3(1-N)u - 10368N4(N-1)2 u3

 

We can, of course, easily transform the problem to finding representations of N as

N =  x3 + y3 + z3 (x + y + z)3

Using the same methodology we find the equivalent elliptic curve to be

FN: v2 = u3 + 36u2 + 432(1-N)u + 1296N2-2592N+1296

with corresponding conditions and transformations.

The most interesting result we found was that the rank-1 curves for F_N have (on average) significantly smaller heights than the E_N curves.

This would seem to be a simple result of the fact that the original problem has 1 cube in the numerator and 3 on the denominator as against 3 cubes in the numerator and only 1 in the denominator for the inverse problem.

The flexibility this allows can be seen from the following parametric solutions for the inverse problem:

(i) x=0 , y=-i , z=i+1 , N=3i²+3i+1
(ii) x=-1 , y=8k+1 , z=-(8k+2) , N=24k²+9k+1
(iii) x=-1 , y=8k+6 , z=-(8k+7) , N=24k²+39k+16