Solutions to the diophantine equation $x^3+y^3+z^3+w^3=1$ What is known about the solutions $(x,y,z,w)\in \mathbb{Z}^4$
of the diophantine equation
$$x^3+y^3+z^3+w^3=1$$
Can you suggest me a book or a paper treating this problem?
 A: Theorem.-The equation $x^3+y^3+z^3+w^3=n$ has an infinity of integer solutions if there exists one solution $(a,b,c,d)$ such that $$-(a+b)(c+d)\gt 0$$ is not a perfect square, and $a\ne b$, or $c\ne d$.
(See Diophantine equations, L. J. Mordell. Academic Press, p. 58).
Since $\mathbb F_7^3=\{0,\pm1\}$ necessarily one of the unknowns is a multiple of $7$ because if not we would have 
$$(\pm1)+(\pm1)+(\pm1)+(\pm1)\equiv 1\pmod7$$ which is impossible.
Trying to find a particular solution I find out $(x,y,z,w)=(14,\space 30,-23,-26)$ which satisfies the conditions of the above theorem. Thus, there are infinitely many solutions which are related, according to the mentioned book, to a Pell's equation (which as it is well known has an infinity of integer solutions).
A: In general it is clear that for such an equations there are infinitely many solutions. For example
$$(n,-n,1,0),\; n\in \mathbb{Z}$$
and permutations. Another, less obvious, possibility is to consider the infinite family of solutions 
$$(9n^4,3n-9n^4,1-9n^3,0)\; n\in \mathbb{Z}$$
It could be interesting to know if all the solutions are contained in a finite number of curves or surfaces.
For a similar problem:
$$x^3+y^3+z^3=1$$
there is a nice discussion in the paper of D.H. Lehmer "On the diophantine equation $x^3+y^3+z^3=1$". I'm searching for an analogous paper in which is considered the case of four variables.
A: This can be solved in pieces:


*

*$a+b+c+d=1$

*$(x)\mapsto x^3$ (used 4 times)

*$\mathbf{Z}^4$ integer lattice


To solve the problem, we need inverse image. The 1st step is just a bar of length 1 split to 4 parts from arbitrary places(in $\mathbf{R}$). in 2nd step, inverse of $x^3$ is just cube root $\sqrt[3]{x}$. 3rd step those cube root values need to hit the lattice points in the integer lattice. So to get solution, we need to understand how cube root behaves. There's a path $$1 \mapsto (a+b+c+d) \mapsto (a,b,c,d) \mapsto (\sqrt[3]{a},\sqrt[3]{b},\sqrt[3]{c},\sqrt[3]{d}) \mapsto P$$ $$ (k_1Z_1+k_2Z_2+k_3Z_3+k_4Z_4) \mapsto P$$, where the first two $\mapsto$ generates infinite tuples satisfying (1), and the last $\mapsto$ filters them away, with $a,b,c,d \in R$ and $k_1,k_2,k_3,k_4 \in Z$. There's a pullback $R^4 \times_P Z^4$.
Note, if you want solutions out from this construct, you should remove the "diophantine" keyword, and thus the lattice disappears, and you can then get solutions from the path. Approximate solution will implement the pullback as just $(x,y,z,t)
 \mapsto (\lfloor{x}\rfloor,\lfloor{y}\rfloor,\lfloor{z}\rfloor,\lfloor{t}\rfloor)$, where there's $\pm 1$ error in the results for each element.
