## 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. 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, so here is where we needed ${0. 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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