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I have a question about the following assertion in this paper, which finds solutions for $33=x^3+y^3+z^3$.

At the bottom of pg 2, it says $$d=|x|-|y|\equiv |z|\mod 3$$ Then it says that for $\epsilon=\{\pm 1\}$, we have

$$x\equiv y\equiv z\equiv \epsilon \mod 3$$

Why is that? It's also possible that $z\equiv 0\mod 3$, and $x\equiv y\mod 3$, right?

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  • $\begingroup$ A few lines before, this was ruled out: "Thus we must have $0 < d < \alpha |z|$." $\endgroup$ – Clement C. Mar 12 at 21:38
  • $\begingroup$ @ClementC.- Does that affect what $d\mod 3$ is? $\endgroup$ – Anju George Mar 12 at 21:39
  • $\begingroup$ Given the rest, I would assume this rules out $z\equiv 0$ (at first glance) $\endgroup$ – Clement C. Mar 12 at 21:50
  • $\begingroup$ @ClementC.- I guess the broader question is why do all three have to be the same $\mod 3$. Another possibility is that $x\equiv -1$, $y\equiv 1$ and $z\equiv 1$ $\endgroup$ – Anju George Mar 12 at 21:53
  • $\begingroup$ Maybe I am just confused, but then you would have $|x|-|y| = 1-1=0\not\equiv |z|=1$. $\endgroup$ – Clement C. Mar 12 at 21:55
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The reason is (partially) given in the part of the sentence which was not quoted:

since every cube is congruent to $0$ or $\pm1 \pmod 9$

More specifically, if $x\equiv a\pmod 3$, where $a$ is $-1,0$ or $1$, then $x^3\equiv a\pmod 9$. This is a very useful restriction, especially if $k\equiv\pm3\pmod 9$.

In particular, if $k = x^3 + y^3 + z^3$ and $k\equiv 3\varepsilon\pmod 9$ with $\varepsilon=\pm1$, the only way to solve the original equation is for all three of $x^3,y^3,z^3$ to be congruent to $\varepsilon\pmod 9$, which means $x,y,z$ are each congruent to $\varepsilon \pmod 3$. Thus we can have $3 \equiv 1 + 1 + 1 \pmod 9$, but it cannot be produced by any other combination of three $-1$s, $0$s and $1$s; similarly for $-3$, which is the relevant value for $k = 33$.

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