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?

  • $\begingroup$ 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

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 $k = 33$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.