The context is the sum-of-four-cubes problem (see here).

I ask myself the following question (I asked a similar question on MSE). Do we know if any integer $k=a^3+b^3+c^3+d^3$ can be decomposed in an infinite number of ways as the sum of four cubes?

I've come up with a very elementary argument (in my opinion, not a new one!) which I think shows that many of the previous $k$ integers can be decomposed in an infinite number of ways.

The idea is to study the diophantine equation $$x^3+(u-x)^3+y^3+(v-y)^3=k\;\;(E),$$ by posing $u=a+b$ and $v=c+d$. This has the particular solution $(x,y)=(a,c)$ and is equivalent to $$(2ux-u^2)^2+uv(2y-v)^2=\frac{u}{3}(4k-u^3-v^3).$$ If we assume that $uv<0$ and $-uv$ is not a perfect square, then I haven't checked the details, but it seems to me that given two integers $x_0$ and $y_0$ such that $(2x_0+1)^2+4uvy_0^2=1$, we can then prove that the generalized Pell's equation $$X^2+uvY^2=\frac{u}{3}(4k-u^3-v^3)$$ has infinitely many solutions $(X,Y)\in\mathbb{Z}^2$ such that $X\equiv-u^2\; (\text{mod }2u)$ and $Y\equiv v\;(\text{mod }2)$, which shows that the equation $(E)$ has an infinite number of solutions $(x,y)\in\mathbb{Z}^2$, and therefore that $k$ can be decomposed as a sum of four cubes in an infinite number of ways.

Do you have any references to this or related questions? Thank you very much!

P.S. Along the way, by trial and error, and using the classic Chakravala method, I found an identity (see $(\star)$ below), surely anecdotal, but which I haven't seen anywhere.

Let $a$, $b$, $c$, $d$, $x_0$ and $y_0$ be any integers.

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 :

$$\begin{equation} \begin{split} A^3+B^3+C^3+D^3-(a^3+b^3+c^3+d^3)&=\\ 3\left(x_0^2+x_0+(a+b)(c+d)y_0^2\right)&\times\left((a-b)^2(a+b)+(c-d)^2(c+d)\right) \end{split} \end{equation}\;\;\;\;(\star)$$

It follows that if $(2x_0+1)^2+4(a+b)(c+d)y_0^2=1$, then $$A^3+B^3+C^3+D^3=a^3+b^3+c^3+d^3.$$

By way of 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 $(2x_0+1)^2+4(a+b)(c+d)y_0^2=1$ with $x_0=288$ and $y_0=51$,

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