Ramanujan's Observation

Once, while visiting the Indian mathematician Ramanujan in the hospital, the British mathematician Hardy remarked that he had arrived in a taxicab whose number, 1729, was quite uninteresting. Ramanujan replied that, on the contrary, 1729 was quite interesting as it was the smallest positive integer which chould be written as the sum of two (positive) cubes in two different ways: 9^3 + 10^3 = 12^3 + 1^3.

We will apply some of the techniques we've learned to investigate the Diophantine equation a^3 + b^3 = c^3 + d^3. As usual, we will let x=a/d, y=b/d, and z=c/d obtaining the equation x^3 + y^3 = z^3 + 1 which we are trying to find rational solutions to. [The set of all real solutions to this equation forms a surface in three-dimensional space.] Note that there are three families of trivial solutions to this equation:

(x,y,z) = (t,1,t)
(x,y,z) = (1,t,t)
(x,y,z) = (t,-t,-1)

Geometrically, these represent three lines lying on the surface we are investigating. We choose one of the lines, say (t,1,t) and consider the family of planes passing through it: y = 1 + p(x - z). If we intersect our surface with this plane, we'll get a cubic curve consisting of a line [corresponding to the line (t,1,t)] and a quadratic curve.

Algebraically:

x^3 + [1 + p(x - z)]^3 = z^3 + 1
(x - z) [x^2 + xz+ z^2 + 3p + 3p^2(x-z) + p^3(x-z)^2] = 0.

The second factor is quadratic in x and z, so if we can find a rational point on it, we can use the techniques from section 1 to find other solutions. Our first attempt to find a rational point might be to look at one of the other lines lying on our cubic surface and see where it meets our intersecting plane. Unfortunately, these lines do not meet the plane in the quadratic, but in the linear component. We try another tack and look at the points where the linear and quadratic components intersect: if we substitute z = x into our quadratic equation, it reduces to

3x^2 + 3p = 0.
This has a solution if p = - q^2 (and then x = z = q). We thus have the quadratic curve

[x^2 + xz+ z^2 - 3q^2 + 3q^4(x-z) - q^6(x-z)^2] = 0
with the rational point (q,q) lying on it. Following the approach of section 1, we take the equation of an arbitrary line through this point

z = m(x - q) + q
and substitute into our quadratic to obtain

-2q^2 - 2m q^2 + m^2 q^2 - 3q^5 + 3m q^5 - q^8 + 2m q^8 - m^2 q^8 + qx + mqx - 2m^2 qx + 3q^4 x - 3m q^4 x + 2q^7 x - 4mx q^7 + 2m^2 q^7 x + x^2 + m x^2 + m^2 x^2 - q^6 x^2 + 2m q^6 x^2 - m^2 q^6 x^2 = 0
which we know must have q as a root and hence the left-hand side has (x - q) as a factor

(x - q)[2q + 2mq - m^2 q + 3q^4 - 3m q^4 + q^7 - 2m q^7 + m^2 q^7 + x + mx + m^2 x - q^6 x + 2mx q^6 - m^2 q^6 x] = 0.
Setting the second factor equal to 0 and solving for x we get

x = (2q + 2mq - m^2 q + 3q^4 - 3m q^4 + q^7 - 2m q^7 + m^2 q^7)/
(-1 - m - m^2 + q^6 - 2m q^6 + m^2 q^6)

z = m(x-q)+q =
(-q + 2mq + 2m^2 q + 3m q^4 - 3m^2 q^4 + q^7 - 2m q^7 + m^2 q^7)/
(-1 - m - m^2 + q^6 - 2m q^6 + m^2 q^6)

y = 1 + p(x-z) = 1 - q^2(x-z) =
(-1 - m - m^2 - 3q^3 + 3 m^2 q^3 - 2 q^6 + 4m q^6 - 2m^2 q^6)/
(-1 - m - m^2 + q^6 - 2m q^6 + m^2 q^6).
Taking m=-1 and q=1/2 we get x=1/10, y=6/5, z=9/10 and recover the solutions a=1, b=12, c=9, d=10. Taking m=1/2 and q=-1/2 we get the new solutions a=219, b=252, c=35, d=298 (and infinitely many others for other choices of m and q).