Hardy's Question

After Ramanujan's comment, Hardy asked what was the smallest positive integer which could be written as the sum of two fourth powers in two different ways. Ramanujan responded that he didn't know, but that he thought it must be quite large. Let's look at the Diophantine equation arising from this question, namely a^4 + b^4 = c^4 + d^4. Letting x=a/d, y=b/d, and z=c/d, we are looking for rational solutions to x^4 + y^4 = z^4 + 1 (a surface). In this case there are eight lines lying on the surface, but we just want to look at two of them:

(x,y,z) = (t,1,t) and
(x,y,z) = (-1,t,t).
We take an arbitrary plane through the first line [y = 1 + p(x - z)] and intersect it with our quartic to get

x^4 + [1+p(x-z)]^4 - z^4 - 1 = 0

which must factor into a linear and a cubic form:

(x-z)[4p + 6p^2 x + 4p^3 x^2 + x^3 + p^4 x^3 - 6p^2 z - 8p^3 xz + x^2 z - 3p^4 x^2 z + 4p^3 z^2 + x z^2 + 3p^4 x z^2 + z^3 - p^4 z^3] = 0.
We want to apply the elliptic curve methods of section 3 to the second factor, so we need a rational point on the curve. If we intersect the line (x,y,z) = (-1,t,t) with the plane y = 1+p(x-z), we find that t = (1-p)/(1+p). This point lies on the cubic factor, rather than the linear one, so we have the elliptic curve

4p + 6p^2 x + 4p^3 x^2 + x^3 + p^4 x^3 - 6p^2 z - 8p^3 xz + x^2 z - 3p^4 x^2 z + 4p^3 z^2 + x z^2 + 3p^4 x z^2 + z^3 - p^4 z^3 = 0
containing the rational point (-1,(1-p)/(1+p)). Finding the equation of the tangent line to this point and substituting into our cubic equation, we get a polynomial which must have (x+1)^2 as a factor. Setting the remaining factor equal to 0 and solving we find that

x = (-1 + 9p + 8p^2 - 6p^3 + 23p^4 - 3p^5 + 2p^6)/
(-2 + 3p - 23p^2 + 6p^3 - 8p^4 - 9p^5 + p^6)

y = (-2 - p - 20p^2 + 17p^3 - 2p^4 + 17p^5 - 8p^6 - p^7)/
((1 + p)(-2 + 3p - 23p^2 + 6p^3 - 8p^4 - 9p^5 + p^6))

z = (1 + 8p - 17p^2 + 2p^3 - 17p^4 + 20p^5 + p^6 + 2p^7)/
((1 + p)(-2 + 3p - 23p^2 + 6p^3 - 8p^4 - 9p^5 + p^6)).
[At this point we should note the advantage of using software packages like Mathematica or Maple for these tedious calculations.]

Taking p = -3 and clearing denominators, we obtain a=133, b=134, c=158, and d=59. A numerical computer check verifies that, in fact, 635318657 = 133^4 + 134^3 = 158^4 + 59^4 is the smallest integer expressible as the sum of two fourth powers in two different ways. [Of course, other values of p will yield more solutions.]