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I have a question as follows and I am pretty sure it is going to come up on my exam

Under what condition on $p$ will a number have a unique cube root $\mod p$, where $p$ is a prime number? Derive an explicit formula for finding cube roots when this condition is met.

So am I correct in saying that p will have a cubed root if $p = 2$ $\mod 3$? I am not sure why this is the case but I remember seeing it somewhere. Whether this is the case or not can someone help me out with this question with a good explanation of the answer? I would really appreciate it

Apurv
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dunika
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  • As there is at least one primitive root use http://math.stackexchange.com/questions/637119/evaluate-the-cubed-root/637135#637135 – lab bhattacharjee Jan 14 '14 at 16:53
  • Do you want to know if every number has a unique cube root mod $p$ or if there exists a number that has a unique cube root mod $p$? – Zur Luria Jan 14 '14 at 16:55
  • @labbhattacharjee would you be able to give more detail as to how I can apply that answer to this question? – dunika Jan 14 '14 at 17:18
  • @user1552404, Now I really fear if you could understand my other solution properly. Now I have added a solution here. Please feel free to get back tome if you have any confusion – lab bhattacharjee Jan 14 '14 at 17:34

2 Answers2

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Hint: express your number as a power of the primitive root. What power(s) can the cube root be? Remember that the order of the multiplicative group mod $p$ is $p-1.$

Igor Rivin
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Let $\displaystyle x^3=a\pmod p\ \ \ \ (1)$

Now as $p$ is prime, $p$ has at least one primitive root, say $g$

Applying Discrete Logarithm wrt $g\pmod p,$

$\displaystyle 3\cdot$ind$_gx\equiv$ind$_ga\pmod{p-1}\ \ \ \ (2)$ which is a Linear Congruence Equation.

Using Linear congruence theorem, $(2)$ hence $(1)$ will be solvable iff ind$_ga$ is divisible by $(3,p-1)=d$. In that case, there will be exactly $d$ solutions

If $p\equiv2\pmod3,p-1\equiv1\implies d=(p-1,3)=1$

lab bhattacharjee
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