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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?

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  • $\begingroup$ Minor nitpick: the double use of $i$ in the line after the displayed equation really makes me wince, and the formulation looks odd: are you sure you mean to say "if $x_i$ belongs to the set $\{x_1,\dots, x_n\}$, then ... " ? $\endgroup$
    – Yemon Choi
    Commented Apr 20, 2015 at 23:45
  • $\begingroup$ I'll edit to address it, thanks. To the question you ask, yes, intended meaning is "if $x$ belongs to the set, then $-x$ does not" $\endgroup$
    – user22139
    Commented Apr 20, 2015 at 23:46
  • $\begingroup$ The "fundamental" condition is not necessary. A minimal solution won't include both $x$ and $-x$. $\endgroup$
    – Zack Wolske
    Commented Apr 20, 2015 at 23:48
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    $\begingroup$ Then the minimum is $N=0$. $\endgroup$
    – Zack Wolske
    Commented Apr 20, 2015 at 23:52
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    $\begingroup$ $N=4$ is not possible even over $\bf C$. The 1st and 3rd elementary symmetric functions of the $x_i$ would vanish, so they would be roots of a polynomial $x^4 + s_2 x^2 + s_4 = 0$, and would thus split into two $\{x,-x\}$ pairs. $\endgroup$
    – Noam D. Elkies
    Commented Apr 21, 2015 at 3:49

4 Answers 4

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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 $2+4+10=7+9$, $2^3+4^3+10^3=7^3+9^3$.

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  • $\begingroup$ Heh, ok, waiting for that final edit where you give an $N=4$ solution or prove it is impossible. $\endgroup$
    – user22139
    Commented Apr 20, 2015 at 23:55
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    $\begingroup$ Obviously, $N=3$ is impossible by Fermat's Last Theorem. $\endgroup$
    – Tony Huynh
    Commented Apr 20, 2015 at 23:56
  • $\begingroup$ I suspect $N=4$ is impossible, if for no other reason than that I can't find an example of it in Gloden's book, Mehrgradige Gleichungen. $\endgroup$
    – Gerry Myerson
    Commented Apr 20, 2015 at 23:58
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    $\begingroup$ If $N=4$ had a solution, then both pairs of (positive) integers would have the same sum and product. Any two numbers are determined by those values, up to ordering. $\endgroup$
    – Zack Wolske
    Commented Apr 21, 2015 at 0:05
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    $\begingroup$ @GerryMyerson That case is impossible since $a^3=(b+c+d)^3 > b^3+c^3+d^3$. $\endgroup$
    – Tony Huynh
    Commented Apr 21, 2015 at 0:11
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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 $N=5$.)

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Suppose there were a solution with $N=4$. Then we have the following two possible cases:

  1. 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.

  2. 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.

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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 $N>4$.

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