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I'm reading an article about numbers that can be expressed as the sum of three cubes:

It states in the article that if you take the following expressions for $x^3+y^3+z^3 = k$ for $x,y,z,k \in \mathbb{N}$ then $k\not\equiv \pm 4\pmod 9$.

I have two questions about this:

  1. It states that a cube modulo 9 always equals 0, -1 or 1. How can you prove this?
  2. Given that this is true, $k \neq 9n \pm 4$ for any integer $n$. I understand how they get to 4, but shouldn't five also be impossible? I get to get 3 modulo 9 you would need 3 cubes that are each 1 modulo nine. For six you would need three cubes of each -1 modulo 9. But how would you reach 5 modulo 9 as the sum of three cubes?

Yes, I am aware another question is about this (If $n\equiv 4 \pmod 9$ then $n$ cannot be written as sum of three cubes?) but this doesn't answer why 5 is possible and why a cube is $0, \pm 1\mod 9$.

Bill Dubuque
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Héctor van den Boorn
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  • If $x$ is divisible by three, then $x^3\equiv 0\pmod 9$ If $x$ is not divisible by $3$ then $x^6\equiv 1\pmod 9$ which means that $(x^3-1)(x^3+1)$ is divisible by $9$. Both terms can't be divisible by $3$ so one of $x^3-1,x^3+1$ must be divisible by $9$. – Thomas Andrews Nov 10 '15 at 21:36
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    $5$ is not possible. $5\equiv -4\pmod 9$. That's why there's a $-$ in the $\pm.$ – Thomas Andrews Nov 10 '15 at 21:37
  • What you need to know in order to answer your question is the Hasse Principal. To find out about it, you can go here $\longrightarrow$ https://m.youtube.com/watch?v=d6nUVa8EtlU – Mr Pie Oct 27 '17 at 08:50

1 Answers1

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  1. Every integer value $x$ is either $3k-1$, $3k$, or $3k+1$ for some integer $k$. If you cube those, you get $27k^3-27k^2+9k-1$, $27k^3$, or $27k^3+27k^2+9k+1$. Each of those is clearly either $-1$, $0$, or $+1$ modulo $9$.

  2. $9k+5 = 9j-4$ for $j = k+1$, and $9k-5 = 9j+4$ for $j = k-1$. So the $\pm 5$ cases are subsumed by the $\pm 4$ cases.

Brian Tung
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  • Thanks for the clear explanation. I'm not a mathematician myself but a programmer and there modulo 4 and modulo 5 are two different things as a modulus is never returned as a negative integer; so that's why I didn't see the equivalence – Héctor van den Boorn Nov 10 '15 at 21:49
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    Specifically, since you're looking at cubes, you could use negative numbers, and thus get negative cubes. and so, not only are you going to look at the likes of 4, but also -4. As you know, and pointed out, "a modulus is never returned as a negative integer" but we're not talking about -4 as the result of taking the mod, but the result you'd get from taking -4, namely 5. By saying it's congruent to ±4 mod 9, you mean the same as saying 4 or 5 mod 9, but with two ("±4") symbols, versus four ("4 or 5") or three ("4,5"). – BMeph Mar 07 '18 at 22:56
  • Easier: by $\mu$LTE: $,n\equiv -1,0,1\pmod{!3},\overset{(\ \ )^{\large 3}}\Longrightarrow, n^3\equiv (-1)^3,0^3,1^3\pmod{!3^2}\ \ $ – Bill Dubuque Apr 15 '24 at 07:53
  • @BillDubuque: Yes. I thought I'd make it more step-by-step for the OP, though. – Brian Tung Apr 15 '24 at 15:34