Infinite number of decompositions into sum of four cubes

Question

The context is the sum-of-four-cubes problem (see here). In this article, it is proved that any integer $k$ of the form $p^3+q^3$ or $2(p^6-q^6)$ (with $p$ and $q$ integers) can be decomposed into the form $$k=P(t)^3+Q(t)^3+R(t)^3+S(t)^3,$$ where $P(t)$, $Q(t)$, $R(t)$ and $S(t)$ are non-constant polynomials with integer coefficients.

It follows that such integers $k$ can be decomposed in an infinite number of ways as a sum of 4 cubes. The author mentions that he doesn't know how to produce such polynomial identities when $k=4$ or $k=5$, for example.

But is it still possible to prove that the preceding values can be decomposed in an infinite number of ways, like the sum of four cubes?

I welcome any references to this or related issues, thank you.

P.S. I posted a similar question on MathOverflow.

Answer 1

Looking more closely at the case $k=4$, a computer program allowed me to conjecture that we had $$4=(x-1)^3+(y+5)^3-x^3-y^3 \;\;(1)$$ for an infinite number of pairs of integers $(x,y)$. The previous equality is equivalent to $$(2x-1)^2-5(2y+5)^2=36.$$ Now it is easy to prove that the Pell's equation $X^2-5Y^2=36$ has an infinite number of solutions $(X,Y)\in\mathbb{Z}^2$ with $X$ and $Y$ odd, it follows that the equation (1) has infinitely many solutions with $x$ and $y$ integers.

Similarly, we can show that we can write $$5=(x+1)^3+(y-2)^3-x^3-y^3,$$ for an infinite number of pairs of integers $(x,y)$. Indeed, the previous equality is equivalent to $$(2x+1)^2-8(y-1)^2=9,$$ and it's easy to prove that the Pell's equation $X^2-8Y^2=9$ has an infinite number of solutions $(X,Y)\in\mathbb{Z}^2$ with $X$ odd.

Without being certain, it seems to me that we can perhaps generalize the preceding reasoning, so as to prove that many integers of the form $k=a^3+b^3+c^3+d^3$ can be decomposed in an infinite number of ways into a sum of four cubes.

Indeed let $u=a+b$ and $v=-(c+d)$ and assume that $uv>0$ and that $uv$ is not a perfect square. Note that $k-u^3+v^3\equiv0\;(\text{mod }3)$.

The equation $$(u-x)^3-(v+y)^3+x^3+y^3=k$$ has the particular solution $(x,y)=(a,c)$, and this is equivalent to $$(2ux-u^2)^2-uv(2y+v)^2=\frac{u}{3}(4k-u^3+v^3),$$ which has the form of a Pell's equation. I haven't looked at the details, but it should be possible to prove that it has infinitely many solutions.

N.B. There may be an interesting identity (see $(\star)$ below) in connection with what I've just said.

Let $a$, $b$, $c$ and $d$ be four integers such that $(a+b)(c+d)<0$ and $-(a+b)(c+d)$ is not a perfect square.

We know that there are then two integers $x$ and $y$ such that $$(2x+1)^2+4(a+b)(c+d)y^2=1.$$

Let $$\left\{\begin{array}{l} A=(a-b)x+(d^2-c^2)y+a\\ B=(b-a)x+(c^2-d^2)y+b\\ C=(c-d)x+(a^2-b^2)y+c\\ D=(d-c)x+(b^2-a^2)y+d\\ \end{array}\right.$$

then $A+B=a+b$, $C+D=c+d$, and $$A^3+B^3+C^3+D^3=a^3+b^3+c^3+d^3\;\;(\star)$$

Here's an example:

We have $a^3+b^3+c^3+d^3=4$ with $a=-5$, $b=1$, $c=4$ and $d=4$,

$(a+b)(c+d)=-32<0$ and $32$ is not a perfect square,

we have $(2x+1)^2+4(a+b)(c+d)y^2=1$ with $x=288$ and $y=51$,

then $A^3+B^3+C^3+D^3=4$ with $A=-1733$, $B=1729$, $C=1228$ and $D=-1220$.

Answer 2

If there is at least a single integer solution to,

$$a^3+b^3+c^3+d^3 = N$$

such that $(a+b)(c+d)<0$ and $(a+b)(c+d) \neq -\square$, then we can find an infinite more using a Pell equation. We use this identity I submitted to Mathworld 20 years ago,

$$A^3+B^3+C^3+D^3 = (a^3+b^3+c^3+d^3)(x^2+wy^2)^3$$

where,

\begin{align} A &= ax^2-v_1xy+bwy^2\\ B &= bx^2+v_1xy+awy^2\\ C &= cx^2+v_2xy+dwy^2\\ D &= dx^2-v_2xy+cwy^2 \end{align}

and,

\begin{align} v_1 &= c^2-d^2\\ v_2 &= a^2-b^2\\ w &= (a+b)(c+d) \end{align}

which is true for any $(x,y)$. Of course, the identity easily proves that $N$ is a sum of four rational cubes in an infinite number of ways. But if we wish to stay in the integers, then we have to require that,

$$x^2+wy^2 = \pm 1$$

thus the requirement $w=(a+b)(c+d)<0$ and $(a+b)(c+d) \neq -\square$, hence a Pell equation. As long as one of the $(a,b,c,d)$ is negative, we can also permute them. For example, we have,

$$1^3+1^3+1^3+1^3 = 4\\ 1^3+4^3+4^3-5^3 = 4$$

We ignore the first solution, and use permutations of the second to get,

$$(1+4)(4-5) = -5\\ (4+4)(1-5) = -32$$

So two Pell equations,

$$x^2-5y^2 = \pm 1\\ x^2-32y^2 = \pm 1$$

allow us to integrally solve $a^3+b^3+c^3+d^3 = 4$ for an infinite number of ways. And similarly for other $N$.