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I wish to find all cubic residues modulo $9$, i.e., all invertible classes $A$ modulo $9$ such that the congruence $X^3 \equiv_{9} A$ is solvable. I have so far determined that the invertible classes are $1, 2, 4, 5, 7, 8$. How do I now best determine those invertible classes for which the congruence is solvable and eliminate those for which it is not?

Evidently, one can just represent any number as $9k+r$, $0 ≤ r < 9$ which, on cubing, gives $729k^3 + 243k^2r+ 27kr^2 + r^3 \equiv_{9} r^3$, and then plug every allowable number for $r$ and obtain its residue..., but I am hoping to find a more efficient way.

Bill Dubuque
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V. Elizabeth
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    Why not just cube each of the residues? – lulu Sep 23 '23 at 15:00
  • @lulu Yes, this is the method I suggested in my post? – V. Elizabeth Sep 23 '23 at 15:07
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    I don't understand. There are only $9$ residues $\pmod 9$. Just compute all their cubes. – lulu Sep 23 '23 at 15:14
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    As a general matter, it's not so easy to decide if something is a cube $\pmod n$ or not. Even if $n=p$ is prime, it's not as simple as it is for quadratic residues. Here, of course, $9$ is so small that you really can do it in your head. – lulu Sep 23 '23 at 15:15

2 Answers2

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We can take your idea about representing any number as $r+9k$ but instead consider every number can be represented as $r+3k$, with $r$ being its residue class mod $3$ so $0\le r \le 2$.

By expanding the binomial all other terms are divisible by $9$ except the first: $(r+3k)^3 = r^3 \mod 9$. So it turns out its residue class mod $3$ determines the cubic residue mod $9$ uniquely! That's simple enough to see by cubing $0,1,2$ we see $0,1,8$ are the only cubic residues mod $9$.

Merosity
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Use the fact that if and only if $r$ is a cubic residue, so is $-r$. So given the set $\{1,2,4,5,7,8\}\equiv\{\pm1,\pm2,\pm4\}$ you need to run trials only on $1,2,4$. These cubes are easy to figure and the only additive-inverse pair you end up with is $\pm1\equiv\{1,8\}$.

Oscar Lanzi
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