This problem is considered in a paper by Andrew Bremner and Richard Guy [bg3].

We first show how to analyse this equation from just this expression.

If we combine the above expression into a single fraction, we must have

-z x2  +  ( N y z - y2 ) x   -  y x2   =   0

and, for this quadratic to have rational solutions, we must have rational solutions of

d2   =   y4  -  2 N z y3  +  N2 z2 y2-  4 z3 y

This quartic has obvious rational solutions so is birationally equivalent to an elliptic curve.

Standard transformation methods show that this curve is

EN :   t2   =   u3  +  N2 u2  +  8 N u  +  16

which can be written as

EN :   t2   =   u3  +  (  N u  +  4  )2

with the transformation

y z  =   N u  -  t  +  4  2 u

leading to

x z  =   -u  4

For example, n = 17 gives the curve  t² = u³ + 289u² + 136u + 16, which has the point

                u = -2664/25,  t = 179284/125.

This leads to  y / z = 1369 / 90  and hence  x / z = 925 / 18, giving   x = 925,  y = 24642,  z = 1620.

The curves have the obvious rational points (0,±4) which are pts of inflexion, and so are of order 3. In general, these and the pt of infinity are the only torsion points, but it is possible that the cubic part of the curve might have a rational root, which is a point of order 2. This occurs only if n=-1,3,5.

Since none of these torsion points give a non-trivial solution of the original problem, we are forced to look for curves with rank > 0.

We can use the Birch and Swinnerton-Dyer conjecture to estimate the rank, and then try to find a rational point of infinite order for those curves with positive rank.

This is significantly more difficult than for most of the other families of curves we look at, because these elliptic curves do NOT have a point of order 2. Such a point makes descent arguments to find a generator much more effective. Bremner and Guy resort to some extremely powerful but complicated computations in cubic fields to find points. These are not for the faint-hearted, and an amateur would find them very difficult. Bremner and Guy supplied solutions for all N in [-99,99], and I extended the search to the range [-199,199].  By March 2003, I had managed to find solutions for most values of N, but there were still a considerable number of values without an answer.

This direct approach can, however, be significantly improved by using the following simple observation. Consider the problem of the equation
 

i3 + j3 + k3 = N i j k

then if  ( i , j , k ) is a solution, then ( x , y , z ) = ( j² k  ,  k² i  ,  i² j ) is a solution to our original problem. Clearly we can generate solutions ( x , y , z ) from much smaller numerical values ( i , j , k ).

We can analyse this relationship in a similar manner to the original form, to show that we need to consider the elliptic curves

IN :  g2 = h3 + N2h2 - 72N h  -  64N3 - 432

with

i / k =  g   - 36  -  N h 6 h + 8 N2

and

j / k =     -  ( g + 36 + N h )  6 h + 8 N2

This approach to the problem has recently (July/August 2003) been analysed in great detail by Dave Rusin who has constructed an excellent set of files on this problem. You should look at the following web-site for much more information:

 www.math.niu.edu/~rusin/papers/research-math/abcn

We can, however, often get a further reduction in the size of the points we need by noting that the curves I_N have a point of order 3 at ( 4N+12 , 4(N² + 3N + 9) ). We can thus do a further 3-isogeny using Velu's formulae to get the curves J_N where

JN :  q2 = p3 + N2p2 - 8 ( 10N3 + 90N2 + 279N + 540 ) p  -

16 ( 4N5 + 92N4 + 616N3 + 2484N2  + 5292N + 6831 )

The transformations from (h,g) to (p,q) are fairly horrendous so I don't write them down, but a computer has no problem coping.

The solutions, to the original problem, can be found in the files brgbpres for N positive, and brgbnres for negative N. Rusin only considers positive N values and has nearly all possible solutions up to N=999.

Latest revision:   8  May  2006.