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 $1 \pmod {p}$
3
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elementary-number-theory
modular-arithmetic
finite-fields
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0Are you trying to ask why the equation $\,y^3=1\pmod p\,$ has 3 solutions, $\,p\,$ a prime? Solutions...in some extension field of $\,\Bbb F_p:=\Bbb Z/p\Bbb Z\,$ , of course...? – 2012-09-20
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1**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? – 2012-09-20
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0$ 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. – 2012-09-20
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0Here we go, $2$ is Gauss, $3$ is Jacobi. – 2012-09-20