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I'm trying to prove the following statement:

Let $p=101$.
Let $xin (mathbbZ/pmathbbZ)^*$ be arbitrary. Show that there exists $yin (mathbbZ/pmathbbZ)^*$ such that $y^3=x$.

Here is my work.
Since $p$ is a prime. We know that the group of units $(mathbbZ/pmathbbZ)^*$ is a field and cyclic.
Let $alphain(mathbbZ/pmathbbZ)^*$ be a primitive root, then $x=alpha^j$ for some $jin mathbbZ$. But I cannot go further.
I also tried Fermat's little theorem as follows; beginequation*
x^100 equiv 1 quad mod(101)
endequation*
and beginequation*
(x^3)^33x equiv 1 quad mod(101)
endequation*
But I cannot see the exact solution. Can you help me please?

Hint: Fermat's little theorem also tells you that $ x^101equiv x mod(101) $, and therefore that $ x^101+100nequiv x mod(101) $ for any integer $ n $. What happens if you choose $ n=1 $?

By the way, the group of units $ (mathbbZ/pmathbbZ)^*$ isn't a field. It's a cyclic group under multiplication, but it's not closed under addition. The entire ring $ mathbbZ/pmathbbZ $ is what constitutes a field.

You almost have it, you just need to flip things around a bit:

$beginequation*
(x^-1)^100 equiv 1 quad mod(101)
endequation*$

$beginequation*
((x^-1)^33)^3(x^-1) equiv 1 quad mod(101)
endequation*$

$beginequation*
((x^-1)^33)^3 equiv x quad mod(101)
endequation*$

So the cube root of $x$ is $(x^-1)^33$.

Consider the map $f: x mapsto x^3$. Since your group is abelian, it is easy to check that $f$ is a group homomorphism. Now, the order of your group is $phi(101) = 100$ and so no non-identity element has an order which is either a divisor or a multiple of 3. So, $textker(f) = x in (mathbbZ/101 mathbbZ)^* = 1$ and hence $f$ is injective. Since $f$ is an injection from a finite set to itself, $f$ is also a surjection and therefore, every element has a pre-image via $f$. That is, every element is a cube.

Hints:

• To prove existence only, prove that the map $x mapsto x^3$ is injective.




• To exhibit a cube root, note that $201=67 cdot 3$.