$Y^3$ congruent to $1 \pmod {p}$ 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 ?).
 A: 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 $p-1$.
A: $y^3-1= (y-1)(y^2 + y + 1)$ 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.
A: 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.
A: 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 $GF(p^2)$.
