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Using p ≡ -1 (mod 3) and p is prime, how can you show $a^3≡b$ (mod p) iff $a≡b^d$ (mod p)? This shows integers mod p has a unique cubic root.

3d ≡ 1 (mod p-1)

I'm not sure where to begin... Does anyone know?

Thanks

omega
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  • What does "integers has a unique cubic root" mean? All the integers or what integers, or where...or what? And what is $;d;$ there? – Timbuc Dec 01 '14 at 04:12
  • You forgot to say what $d$ is. Which part are you having trouble with? Showing the "iff" part? Or showing that it implies uniqueness of cube root? Or something else? –  Dec 01 '14 at 04:12
  • Pretty much all of it. – omega Dec 01 '14 at 04:13
  • Usually, I find that when one is bewildered by a problem, one is trying to take in too much at once. Try breaking it apart into smaller pieces. –  Dec 01 '14 at 04:15
  • Working out some example(s) may help, say with $;p=5,,,11;$ and etc. – Timbuc Dec 01 '14 at 04:16
  • And I still cannot understand the role of that $;d;$ there. It is supposed to be "some integer" s.t. $;3d=1\pmod{p-1};$ ? – Timbuc Dec 01 '14 at 04:18
  • I also don't understand the role, but its supposed to be used to show the proof. I know that d will always exist for any p. – omega Dec 01 '14 at 04:21
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    One direction, $b^d\equiv a^{3d}=a^{(k)(p-1)+1}\equiv a^1\pmod{p}$. The question will probably be closed while I try to type the full answer, so you will have to take care of the other direction. Not hard. – André Nicolas Dec 01 '14 at 06:11
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    There is by the way a more natural way to prove the ultimately desired result. Show the cube function mod $p$ is one to one by showing that if $a^3\equiv a_1^3$ then $a\equiv a_1$. Let $c_1$ be the inerse of $a_1$. Then $(ac_1)^3\equiv 1$, so the order of $ac_1$ is $3$ or $1$. But $3$ does not divide $p-1$. – André Nicolas Dec 01 '14 at 06:23
  • Possible duplicate of Quotient ring of Gaussian integers – Caleb Stanford Sep 01 '16 at 20:12

1 Answers1

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Let $p$ be a prime that is congruent to $-1$ modulo $3$. Then $3$ does not divide $p-1$. It follows that there is a positive integer $d$ such that $$3d\equiv 1\pmod{p-1}.\tag{1}$$

We show that for any $a$ and $b$, and any positive $d$ that satisfies (1), we have $$a^3\equiv b\pmod{p}\quad\text{if and only if}\quad a\equiv b^d\pmod{d}.$$ The result is obvious if one of $a$ or $b$ is divisible by $p$. So we may assume that $a$ and $b$ are each relatively prime to $p$.

Suppose that $a^3\equiv b\pmod{p}$. Then $(a^3)^d\equiv b^d\pmod{p}$. But $3d\equiv 1\pmod{p-1}$. So $3d=(p-1)k+1$ for some integer $k$. It follows that $$b^d\equiv a^{3d}=(a^{p-1})^k a^1\equiv a\pmod{p}.$$ In the above calculation, we used Fermat's Theorem to show that $a^{p-1}\equiv 1\pmod{p}$.

For the other direction, suppose that $a\equiv b^d\pmod p$. We want to show that $a^3\equiv b\pmod{p}$. We use the fact that $3d$ is of the shape $(p-1)k+1$. We have $$a^3\equiv b^{3d}=(b^{p-1})^kb^1\equiv b\pmod{p}.$$

Remark: If $p\equiv -1\pmod{3}$, a suitable $d$ is quickly computed, even for large $p$, by using the Euclidean Algorithm. Now given any $b$, we can compute cheaply the modular cube root $a$ of $b$ by calculating the remainder when $b^d$ is divided by $p$. This too can be done efficiently, by using the "binary" method for exponentiation.

So not only does the modular cube root of $b$ exist, it can also be efficiently computed. The story is more messy if $p\equiv 1\pmod{3}$. In that case, only one-third of the numbers is the interval $1\le b\le p-1$ have modular cube roots, and finding one, when it exists, is computationally more complicated for large $p$.

André Nicolas
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  • I recognize this is old, but why does it follow in line 2 that such a positive integer d exists? I can see why one would exist with 3d≡2 mod p-1, but not one with 3d≡1 – MathTrain Apr 25 '19 at 07:27