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How to get the condition on $p$ for which $y^3$ congruent to $1$ modulo $p$ has $3$ solutions ( $1$ solution $x= 1$ is always possible, right ?).

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Are you trying to ask why the equation $\,y^3=1\pmod p\,$ has 3 solutions, $\,p\,$ a prime? some extension field of $\,\Bbb F_p:=\Bbb Z/p\Bbb Z\,$ , of course...? – DonAntonio Sep 20 '12 at 17:46
Hint: $\mathbb{F}_p$ is a field, and $y^3 - 1 = 0$ if and only if $(y-1)(y^2 + y + 1) = 0$. So the question is: for what $p$ does $y^2 + y + 1 = 0$ have two solutions? – JavaMan Sep 20 '12 at 18:19
$ y^3 \equiv 2 \pmod p $ is more interesting. $ y^3 \equiv 3 \pmod p $ is at the same level of difficulty, you don't see it quite as often. – Will Jagy Sep 20 '12 at 21:00
Here we go, $2$ is Gauss, $3$ is Jacobi. – Will Jagy Sep 20 '12 at 21:09

Let $\mathbb{F}^\ast$ denote the group of nonzero elements of the finite field order $p$.

Clearly the roots of $y^3-1$ form a subgroup of $\mathbb{F}^\ast$ of order 3 or 1, and any such subgroup of order three has the roots of $y^3-1$.

That abelian group has a subgroup of order 3 iff $3$ divides $|\mathbb{F}^\ast|=p-1$.

A problem that can happen is that if the characteristic of $\mathbb{F}$ is 3, then then $y^3-1=(y-1)^3$ only has 1 distinct root. I'm not positive, but I'm pretty sure as long as $p$ is not a power of 3, it always has distinct roots.

Of course, $p$ would not be a power of 3 if 3 divided $p-1$.

So! I believe the criterion is that 3 divides $p-1$.

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If it is a finite field, then $p$ is a prime. Except $3$ itself, no prime is a power of $3$. – celtschk Sep 20 '12 at 21:21
@celtschk You'll notice that I never assumed $p$ was prime (just that $\mathbb{F}$ is a field of order $p$), so that is a non-issue. Since the OP was unclear on the point, I left the order of the field open to be a power of a prime. Yeah, $p$ suggests "prime", but there is no reason to restrict to that case. – rschwieb Sep 21 '12 at 12:15

$y^3-1= (y-1)(y^2 + y + 1)$ and so $y^3-1=0$ has more roots iff $y^2 + y + 1 = 0$ has a root $y\ne 1$. By the classical quadratic formula, this happens iff $-3$ is a square mod $p$. This happens exactly when $p=3$ or $p\equiv 1 \bmod 3$, by quadratic reciprocity.

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Something's off with the first line, because $y^2+y+1=0$ has 1 as a root in $\mathbb{F}_3$, but 1 is the only root of $y^3-1$ over $\mathbb{F}_3$. – rschwieb Sep 20 '12 at 19:33
@rschwieb, you're right, $p=3$ is a special case. – lhf Sep 20 '12 at 20:40
I added explension for why this is a special case – Belgi Sep 20 '12 at 20:49
But for $p=3$, the only non-zero elements are $1$ and $2$, so there's no way you get 3 different roots. – celtschk Sep 20 '12 at 21:25

To add to the answers given and address the question about when all the solutions are different: we are looking at the polynomial $x^{3}-1$ over the field $\mathbb{F}_{p}$.

The question is when this polynomial have distinct roots (in some extension): if $p=3$ then $(x^{3}-1)'=3x^{2}\equiv0$ so its not separable and have some multiple roots.

When $p\neq3$ we have $(x^{3}-1)'=3x^{2}\not\equiv0$ hence have $3$ distinct roots.

So this is the reason $p=3$ is a special case. about when all the roots are in $\mathbb{F}_{p}$it self is in the other answers.

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The group $\mathbb{Z}/p\mathbb{Z}^*$ is cyclic of order $p-1$ (look up primitive element/root). The question is about the existence of elements of order 3 (the element of order 1 is always there). The basic theory of cyclic groups tells that these exist, iff $3\mid p-1$ or, equivalently, $p\equiv1\pmod3$.

If $p$ does not satisfy this congruence, then the primitive cubic roots of unity reside in the (unique up to isomorphism) quadratic extension field $GF(p^2)$.

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