The quartic surfaces x⁴ + y⁴ + z⁴ = N

The most famous quartic surface of this kind is Euler's, the case where N=1, and its fame arises from the fact that, after Euler considered it in 1769, no points were found on it until work of Noam Elkies in 1986. The family is in some sense the obvious extension of the Fermat curves x³ + y³ = N; the problem is that, whilst the Fermat curves are elliptic curves and so can be handled using the full panoply of arithmetic and computational methods invented by people from Weierstrass in the 1890s to Cremona in the 21st century, the algebraic structure of quartic surfaces is a lot more involved and a lot less well-understood.

That said, I'll take my shovel and my bludgeon and start wading in.

Software

There are two obvious and reasonable efficient approaches to finding points on a quartic surface, and one less-obvious and significantly more efficient one. You can construct pairs [x,y, x⁴+y⁴] in one priority queue, one pair per value of x, and pairs [z,y, Nt⁴-z⁴] in another, one pair per value of t, then interleave the two priority queues and look for shared values of the sum.

You can store pairs [x,y] in a hash table keyed by the value of x⁴ + y⁴, then enumerate sums Nt⁴ - z⁴ and see if they match any keys in the hash table ; this avoids the overhead of sorting, and seems to be noticably faster than the priority-queue approach. On a modern PC it takes an hour or so for a search bound of 100,000.

The clever method is to pick a medium-sized prime p (where 'medium-sized' means you're prepared to compute p^2 fourth roots and perform p^2 4x4 lattice reductions), then search over u and v for modulo-p solutions to 1^4 + u^4 + v^4 - N*t^4 = 0. For each such solution, you look at the points modulo p^2 which reduce to it modulo p and which satisfy the equation mod p^2, and indeed one condition modulo p^3. These turn out to be linear conditions, and to give you a lattice of determinant p^5; look at the small points in this lattice and see which of them satisfy the condition over the integers. This algorithm is due to Noam Elkies, and was published in 2000 in the Proceedings of the ANTS-4 conference; Mark Watkins has implemented this algorithm in the most recent version of the magma computer algebra package.

Easy p-adic constraints

All fourth powers are equal to 0 or 1 modulo 5; all fourth powers are equal to 0 or 1 modulo 16. So N, being a sum of three fourth powers, must equal 1, 2 or 3 modulo both 5 and 16 - it clearly can't be four or more, and if it's zero then a fairly standard descent argument shows that there can be no integer solutions.

If a^4+b^4+c^4 = N*d^4 with N=1 mod 5, then precisely one of a, b and c must not be a multiple of five; we can require a and b to be multiples of 5 and c and d not. If N=2 mod 5 then precisely one of a, b and c must be a multiple of five; we require it to be c. If N=3 mod 5 then a, b and c must all be non-multiples of five. This makes the enumerative searches mentioned in the previous section a good deal faster.

A very similar rule works for the different congruence classes of N modulo 16; d must always be odd, and one, two and three of a, b and c must be odd, depending on whether N is 1, 2 or 3 mod 16.

Points that I know for N < 500

The points that I've found are on an automatically-generated page here. Most of the longer calculations were done over the August Bank Holiday weekend (87 hours, if you count from 6pm on the Friday to 9am on the Monday) on the idle cycles of a few dual-core 2200MHz Athlon64 workstations; N=163 was done over that weekend on a 2.4Ghz Pentium 4.

Infinite families of points

Given by a hyperplane section

u⁴ + v⁴ + (u+v)⁴ factorises as 2(u² + uv + v²)². f(u,v) = u² + uv + v² is a quadratic form with a nice arithmetic structure; in particular, if X can be written as f(u1,v1), X² can be written as f(v²-u², u(u+2v)). So if you have any number that can be written in the form f(u,v), which is any product whose prime factors are all either equal to 3 or equal to 1 mod 6, you can find a point on u⁴ + v⁴ + w⁴ = 2 with that number as the denominator.

For example, 666667 = 378^2 + 378*559 + 559^2, so
666667^2 = 169597^2 + 169597*565488 + 565488^2, and so
169597^4 + 565488^4 + 735085^4 = 2 * 666667^4.

This family is so enormous that I have deliberately not listed any points from it on the page of known points; they'd crowd out the more interesting 'sporadic' ones.

Given by an embedded elliptic curve

The quartic surface fairly obviously lives in P³, since it's given by a homogenous equation in 3+1 variables. In P³, an intersection of two quadrics gives an elliptic curve; and good machinery exists for handling points on elliptic curves.

Noam Elkies was able to find an entire two-dimensional family of embedded elliptic curves by pure thought; I, on the other hand, cheat. Having found some points with reasonably large coefficients by naive search, I use lattice reduction to find conics in three or four variables, with small coefficients, which pass through one of the points, then ask magma to iterate over pairs of such conics until it finds a pair Q1, Q2 such that the ideal generated by Q1 and Q2 contains the equation of the quartic surface.

I've found three curves so far, after getting magma to consider a few tens of millions of pairs of conics:

N=1	C1:= 13*x^2 - 21*x*y - 5*x*t + 13*y^2 + 5*y*t - 8*z^2 - 8*t^2 C2:= 8*x^2 - 5*x*y + 21*x*t + 8*y^2 - 21*y*t + 13*z^2 + 3*t^2
N=2	C1:=49*x^2 + 370*x*z - 257*y^2 + 49*z^2 C2:=185*x^2 + 49*x*z + 185*z^2 - 257*t^2
N=18	C1:= 17*x^2 + 7*y^2 - 26*y*z + 7*z^2 C2:=13*y^2 - 7*y*z + 13*z^2 - 51*t^2

Each of these pairs of quadrics corresponds to an elliptic curve (of rank two for the N=1 and N=18 examples, of rank one and with quite a large generator for N=2), and the construction is such that all but finitely many rational points on that curve map back to the quartic surface. Generally, the torsion of the elliptic curve corresponds to some of the obvious symmetries (permuting X, Y and Z, or negating any of them) of the surface, and the generators of the curve give two dimensions-worth of points on the surface.

The magma routine here finds points on the N=18 surface using the second of these curves; whilst I found the curve by noting that it went through the point [6617, 18454, 61247, 29797], it happens also to go through [1,1,2,1]. The smaller points generated by the curve are listed here; if you compare with the list of points found by enumeration, you can see that there's probably another curve to be found.

The curve on the N=2 surface hits only the first of the points found by enumeration, and next encounters the unreasonably large point [4014369822293252845298556668977, 8028813922964684804294250901215, 8331871536210073175631303374584, 8243128117136361914922521992201]

The searches for points and for curves continue; if you're interested in contributing brain or CPU, mail me.