The context is the sum-of-four-cubes problem (see [here][1]).

I ask myself the following question (I asked a similar question on [MSE][2]). 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!

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**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 

$$\left\{\begin{array}{l}
A=(a-b)x_0+(d^2-c^2)y_0+a\\
B=(b-a)x_0+(c^2-d^2)y_0+b\\
C=(c-d)x_0+(a^2-b^2)y_0+c\\
D=(d-c)x_0+(b^2-a^2)y_0+d\\
\end{array}\right.$$

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 $D=-1220$.




  [1]: https://en.wikipedia.org/wiki/Sum_of_four_cubes_problem
  [2]: https://math.stackexchange.com/questions/4968479/infinite-number-of-decompositions-into-sum-of-four-cubes