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Let $p,q$ be two odd prime numbers such that $3\mid(p-1)$ and $3\nmid(q-1)$.

I need to show that:
a) $1$ has $3$ cube roots modulo $p$.
b) If $a,b\in(\mathbb{Z}/q\mathbb{Z})^*$ such that $b^3\equiv a\pmod{q}$, then $b$ is the only cube root of $a$ modulo $q$.

I tried to use Fermat's Theorem and some manipulations on the equations but I can't figure how to prove that.

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  • 1
    $\begingroup$ A quick answer to (b): since $3\nmid(q-1)$, there exists $y$ such that $3y\equiv1\pmod{q-1}$; say $3y=1+t(q-1)$. Then $b^3\equiv a\pmod q$ implies $a^y\equiv b^{3y}=b\cdot (b^{q-1})^t \equiv 1\pmod q$ by Fermat's little theorem (since $b\in(\Bbb Z/q\Bbb Z)^*$). In other words, any solution $b$ of $b^3\equiv a\pmod q$ must satisfy $b\equiv a^y\pmod q$, which proves uniqueness. $\endgroup$ – Greg Martin Jan 3 '15 at 23:23
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  • For any prime number $\ell$, the group of units $(\mathbb{Z}/\ell\mathbb{Z})^*$ is a cyclic group of order $\ell-1$, i.e. $$(\mathbb{Z}/\ell\mathbb{Z})^*\cong\mathbb{Z}/(\ell-1)\mathbb{Z}$$ Note that the operation is multiplication for the first group, and addition for the second group. Explicitly, if $g\in(\mathbb{Z}/\ell\mathbb{Z})^*$ is any primitive root modulo $p$, then the function $$\varphi:\mathbb{Z}/(\ell-1)\mathbb{Z}\longrightarrow(\mathbb{Z}/\ell\mathbb{Z})^*,\quad \varphi(t)=g^t$$ is an isomorphism.

  • To say that $a\in(\mathbb{Z}/\ell\mathbb{Z})^*$ is a $k$th root of unity, is the same as to say that $a^k=1$, which is the same as to say that the order of $a$ in the group $(\mathbb{Z}/\ell\mathbb{Z})^*$ divides the number $k$.

  • If $f:A\to B$ is any isomorphism of groups, then $f$ preserves the order of elements. In other words, for any $a\in A$, the order of $f(a)$ in $B$ is equal to the order of $a$ in $A$, and for any $b\in B$, the order of $f^{-1}(b)$ in $A$ is equal to the order of $b$ in $B$.

  • Now, you should figure out for yourself: if $A$ is a finite cyclic group, say of order $n$, how many elements of order $d$ are there in $A$?

  • You should also figure out for yourself: if $A$ is any abelian group, with $a,b,c\in A$ and $b^3=a$ and $c^3=a$, what is $(bc^{-1})^3$? If you then require that $1$ is the only cube root of unity in $A$, what does your observation about $(bc^{-1})^3$ tell you?

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