# A question about the famous paper that finds solutions for $33=x^3+y^3+z^3$

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?

• A few lines before, this was ruled out: "Thus we must have $0 < d < \alpha |z|$." – Clement C. Mar 12 at 21:38
• @ClementC.- Does that affect what $d\mod 3$ is? – Anju George Mar 12 at 21:39
• Given the rest, I would assume this rules out $z\equiv 0$ (at first glance) – Clement C. Mar 12 at 21:50
• @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 $z\equiv 1$ – Anju George Mar 12 at 21:53
• Maybe I am just confused, but then you would have $|x|-|y| = 1-1=0\not\equiv |z|=1$. – Clement C. Mar 12 at 21:55

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 $$k = 33$$.