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 ?).
$Y^3$ congruent to $1 \pmod {p}$
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0Here we go, $2$ is Gauss, $3$ is Jacobi. – 2012-09-20
4 Answers
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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0@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*. – 2012-09-21
$y^3-1= (y-1)(y^2 + y + 1)$ 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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0But for $p=3$, the only non-zero elements are $1$ and $2$, so there's no way you get 3 different roots. – 2012-09-20
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.
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 $GF(p^2)$.