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If $p$ is congruent to $2 \pmod 3$, how can I prove that all $1 \le a \le p-1$ are cubic residues $\bmod p$?

Here's what I've done:

$1^3$ congruent to $1 \pmod p$ thus, 1 is a cubic residue. Also $(p-1)^3=(p-1) \pmod p$ implies $$(p-1)^3 - (p-1) = (p-1)[(p-1)^2 -1] = (p-1)(p^2 -2p) = p(p-1)(p-2) $$ implies $$p\mid(p-1)^3 -(p-1).$$ Thus $p-1$ is a cubic residue $\bmod p$. Now I don't know how to show for $1\lt a \lt p-1$ are all cubic residues when $p$ is congruent to $2\pmod 3$.

user26857
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Gaby
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4 Answers4

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Hint: Every $1\leq a\leq p-1$ is a cubic residue if and only if the cube map $c:(\mathbb{Z}/p\mathbb{Z})^\times\to (\mathbb{Z}/p\mathbb{Z})^\times$, defined by $c(x)=x^3$, is surjective. A function from a finite set to itself is surjective if and only if it is injective, and a homomorphism is injective its kernel is trivial. The kernel of $c$ is $$\ker(c)=\{x\in(\mathbb{Z}/p\mathbb{Z})^\times\mid x^3=1\}.$$ Now think about Lagrange's theorem, the possible sizes of $\ker(c)$, and the size of $(\mathbb{Z}/p\mathbb{Z})^\times$ given that $p\equiv 2\bmod 3$.

Zev Chonoles
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    Can you explain this in a strictly number theory view, without algebraic structures? – Gaby Jun 20 '13 at 19:54
  • I very much doubt there's such a thing, @Gaby...at least in a general case. – DonAntonio Jun 20 '13 at 19:58
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    I don't know any other way of doing it (of course there may still be one). Also, basic abstract algebra unifies and simplifies so much of basic number theory - I think it is extremely worthwhile learning to think in this way. – Zev Chonoles Jun 20 '13 at 20:00
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    Indeed, if you rephrase the question as, "Show that the map $x\to x^{3}$ is an automorphism on $(\mathbb Z / p\mathbb Z)^{\times}$, when $p \equiv 2 \pmod{3}$" then it is more clear to approach this problem algebraically. – Dylan Yott Jun 20 '13 at 20:10
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    Admittedly, this isn't just a rephrasing, it requires a little work (which Zev mentioned) to show that this is the same as the problem. – Dylan Yott Jun 20 '13 at 20:34
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We use more machinery than necessary.

Let $g$ be a primitive root of $p$. Then for some $k$ we have $a\equiv g^k\pmod{p}$.

Let $p=3n+2$. Then by Fermat's Theorem we have $g^{3n+1}\equiv 1\pmod{p}$, and $g^{6n+2}\equiv 1\pmod{p}$.

Note that $$a\equiv g^k\equiv g^{k+3n+1}\equiv g^{k+6n+2}\pmod{p}.$$ Exactly one of $k$, $k+3n+1$, and $k+6n+2$ is divisible by $3$. Call it $3d$. Then $$a \equiv (g^d)^3 \pmod{p},$$ so we have obtained a semi-explicit expression for the cube root of $a$ modulo $p$.

user26857
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André Nicolas
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Claim: Suppose $p$ is a prime congruent to $2\bmod 3$. Then, $0^3,1^3,2^3,\ldots,(p-1)^3$ are all different modulo $p$.

Hint for the proof of the claim: The result is clear if $p=2$, so assume $p>2$. Suppose $u^3\equiv v^3 \bmod p$. Then $u\equiv v \bmod p$, or $u^2+uv+v^2\equiv 0 \bmod p$. But the latter is impossible when $p\equiv 2 \bmod 3$ (complete the square).

Corollary: There is an equality of sets $\{0^3,1^3,2^3,\ldots,(p-1)^3\}=\{0,1,2,3,\ldots,p-1\}$, i.e., every congruence class in $\mathbb{Z}/p\mathbb{Z}$ is a cube.

Álvaro Lozano-Robledo
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    When you say the latter is impossible, my first instinct would be to fix $v \neq 0$, then look at the discriminant and use QR, but this seems harder than it should be. – Dylan Yott Jun 20 '13 at 20:29
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Let $x^3\equiv a\bmod p$. We also have $a^{p-1}\equiv 1$, so:

$a^{2p-1}\equiv (a^{p-1})^2a\equiv a\bmod p$

but also $2p-1\equiv0\bmod3$, thus $2p-1=3k$ for some whole number $k$.

Thereby

$x^3\equiv a\equiv a^{3k}\bmod p,$

from which $x\equiv a^k$ is a guaranteed cube root of $a$.

Oscar Lanzi
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