Nontrivial solutions for $\sum x_i = \sum x_i^3 = 0$ For $x_i \in \mathbb{Z}$, let  $\{x_i\}$ be a fundamental solution to the equations:
$$
\sum_{i= 1}^N x_i = \sum_{i=1}^N x_i^3 = 0
$$
if $x \in \{x_i\} \Rightarrow -x \notin \{x_i\}$.
For instance, a fundamental solution with $N=7$ is given by
$$
x_1 = 4, \quad  x_2 = x_3 = x_4 = -3, \quad  x_5 = x_6 = 2, \quad  x_7 = 1
$$
What is the minimum $N$ for which a fundamental solution exists?
 A: Suppose there were a solution with $N=4$. Then we have the following two possible cases:


*

*Three of the $x_i$ have the same sign. Then w.l.o.g. we have $x_1,x_2,x_3>0$ and $(x_1+x_2+x_3)^3=x_1^3+x_2^3+x_3^3$, which is impossible.

*Two of the $x_i$, say $x_1$ and $x_2$ are positive, and $-x_3$ and $-x_4$ are positive. Then we have
$x_1+x_2=x_3+x_4$ and $x_1^3+x_2^3=x_3^3+x_4^3$. We can deduce from these equations that also $x_1^2+x_2^2=x_3^2+x_4^2$. Now pick $\alpha$ such that $x_3=x_1+\alpha$, $x_4=x_2-\alpha$. It follows that $2\alpha(x_1-x_2)+2\alpha^2=0$, so either $\alpha=0$ or $\alpha=x_2-x_1$, but these cases are excluded. 
A: If $N=6$ then we have a nice identity:
$$(x_1+x_2)(x_1+x_3)(x_2+x_3)=-(x_4+x_5)(x_4+x_6)(x_5+x_6).$$
(Loo-Keng Hua used such identities in his "Additive Theory of Prime Numbers".) In particular if $x_5=x_6=0$ then LHS vanishes. From this observation easily follows that $N>4$.
A: For $N=4$ we get the projective cubic curve
$$
x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3.
$$
But this is just the union of $x_1=-x_2$, $x_1=-x_3$, and $x_2=-x_3$, contrary to your requirements. Hence $N \geq 5$. Therefore Gerry Myerson's solution is optimal. 
Another way to see that $N=5$ can be attained is as follows: for $N=5$, the equations define the smooth, projective cubic surface $X$ given by
$$
x_1^3+x_2^3+x_3^3+x_4^3=(x_1+x_2+x_3+x_4)^3,
$$
called the Clebsch surface by some authors. Now since $X$ has a rational point, e.g. $(0:0:0:1)$, it has a Zariski dense set of them (for example by Segre--Manin), and therefore it has a rational point $(x_1:x_2:x_3:x_4)$ with all $x_i$ integral and $x_i \neq -x_j$ for $i \neq j$. (Admittedly this argument is phrased in a non-constructive way, but it is in fact easy to write down a birational parametrization of $X$, and so construct a plethora of integral solutions to the system with $N=5$.) 
A: We can get $N=6$ from $1+5+5=2+3+6$, $1^3+5^3+5^3=2^3+3^3+6^3$. 
We can get $N=5$ from $2+4+10=7+9$, $2^3+4^3+10^3=7^3+9^3$. 
