Every large number is the sum of 13 cubes

An important function that arises from Waring’s problem is the the following. For any {k}, let {G(k)} be the least {s} such that, for all {n} sufficiently large (how large can depend on {k} and {s}) there exist positive integers {n_1,\hdots,n_s} such that

\displaystyle  n=n_1^k+\cdots+n_s^k.

Waring’s problem itself says that {G(k)} actually exists for all {k\geq 2}. In other words, past some (possible very large) integer, all integers are the sum of at most some constant number of {k}th powers. The classical case is Lagrange’s theorem, which states that {G(2)=4}. In fact, we can make the stronger statement that all integers, not just the large ones, are the sum of at most four squares.

In this post, I’ll give a nice bound for {G(3)} (and hence, in particular, {G(3)} actually exists). I’ll be following section 21.2 of Hardy and Wright.

Theorem 1

\displaystyle G(3)\leq 13.

In what follows, {z} will always denote an integer congruent to 1 modulo 6. Define

\displaystyle \phi(z)=11z^9+(z^3+1)^3+125z^3=12z^9+3z^6+128z^3+1,

and

\displaystyle \psi(z)=14z^9.

For large {z} we have {\phi(z+6)<\psi(z)} — a quick computation shows that this is true for {z\geq 115}, for example. It follows that if we define the interval {I_z=[\phi(z),\psi(z)]} then these overlap, and so for {z\geq 115} these intervals overlap. In particular, every {n\geq 10^{20}>14(115^9)} is in one of these intervals.

It suffices to show, therefore, that if {n\in I_z} for some fixed {z} then we can write {n} as the sum of at most {13} cubes.

Let {r}, {s}, and {N} be the smallest positive integers such that

\displaystyle n\equiv 6r\pmod{z^3},

\displaystyle n\equiv s+4\pmod{6},

and

\displaystyle N=(r+1)^5+(r-1)^3+2(z^3-r)^3+(sz)^3.

In particular, note that {N} is the sum of 5 cubes. Furthermore,

\displaystyle 8z^9<\phi(z)-N\leq n-N<14z^9.

Also,

\displaystyle N\equiv (r+1)^3-(r-1)^3-2r^3\equiv 6r\equiv n\equiv n-8z^9\pmod{z^3}.

Similarly, {N\equiv n-8z^9\pmod{6}}. Hence {n-N-8z^9} is a multiple of {6z^3}, so we can write

\displaystyle n=N+8z^9+6mz^3

for some {0<m<z^6}. If {m} were some constant, we could stop here having shown that {G(3)\leq 12}, but unfortunately {m} could be increasing with {z}, and hence {n}, so a further reduction is required.

Here we use the fact mentioned above, that {G(2)\leq 4}. Hence we can write {m=x_1^2+x_2^2+x_3^2+x_4^2}. Plugging this into the above and doing some elementary algebra proves the identity

\displaystyle n=N+\sum_{i=1}^4\left( (z^3+x_i)^3+(z^3-x_i)^3\right).

We have written {n} as the sum of 13 cubes, and all that remains is to check that these are positive cubes — this follows from the fact that {0\leq x_i<z^3}, so here is where we needed {0<m<z^6}. The proof is complete.

A couple of final notes:

1) By checking all {n<10^{20}} by hand, or by being clever first and then checking by hand, it can be shown that in fact every integer is the sum of at most 13 cubes.

2) The right answer is {G(3)=9}. This can be proved in a way similar to the above if we assumed that every integer which is not of the form {4^a(8m+7)} is actually the sum of three squares. This is true, but difficult.

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